Datasets:
ericsun153
/
mathqa_question_shift_test

Dataset card Data Studio Files
xet
Community
Problem
stringlengths
5
628
Rationale
stringlengths
1
2.74k
options
stringlengths
37
137
correct
stringclasses
5 values
annotated_formula
stringlengths
6
848
linear_formula
stringlengths
7
357
category
stringclasses
6 values
mr . shah decided to walk down the escalator of a tube station . he found â that if he walks down 26 steps , he requires 30 seconds to reach the bottom . however , if he steps down 34 stairs he would only require 18 seconds to get to the bottom . if the time is measured from the moment the top step begins â to descend to the time he steps off the last step at the bottom , find out the height of the stair way in steps ?
"explanation : age of the 18 th student = [ 18 * 18 - ( 14 * 5 + 16 * 9 ) ] = ( 324 - 214 ) = 110 years . answer : c"
a ) 101 , b ) 66 , c ) 110 , d ) 160 , e ) 12
c
subtract(multiply(18, 18), add(multiply(5, 14), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
the average age of 18 students of a class is 18 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 18 th student is
"sol . 15 % of rs . 34 = rs . ( 15 / 100 x 34 ) = rs . 5.10 answer d"
a ) rs . 3.40 , b ) rs . 3.75 , c ) rs . 4.50 , d ) rs . 5.10 , e ) none
d
divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
what is 15 percent of rs . 34 ?
"d = ( d - r ) / q = ( 123 - 4 ) / 7 = 119 / 7 = 17 c"
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19
c
floor(divide(123, 7))
divide(n0,n1)|floor(#0)|
general
on dividing 123 by a number , the quotient is 7 and the remainder is 4 . find the divisor .
solution : 1 st method : let the original price of tea be rs . x / kg . after reduction the price becomes = x - 10 % of x = 9 x / 10 per kg . now , ( 22500 / ( 9 x / 10 ) ) - 22500 / x = 25 or , 22500 [ 10 / 9 x - 1 / x ] = 25 or , 25 * 9 x = 22500 ; or , x = ( 22500 / 2589 ) = rs . 100 . hence , new price = 90 per kg . thought process method : let the original price be rs . 100 per kg , he get tea = 22500 / 100 = 225 kg . after reduction the price becomes = 90 per kg , he get tea = 22500 / 90 = 250 kg . so , reduction price is rs . 90 per kg as it enables him to buy 25 kg of more tea . answer : option c
a ) rs . 70 , b ) rs . 80 , c ) rs . 90 , d ) rs . 100 , e ) none
c
divide(divide(multiply(22500, subtract(divide(const_100, subtract(const_100, 10)), const_1)), 25), divide(const_100, subtract(const_100, 10)))
subtract(const_100,n0)|divide(const_100,#0)|subtract(#1,const_1)|multiply(n2,#2)|divide(#3,n1)|divide(#4,#1)
gain
a reduction of 10 % in the price of tea enables a dealer to purchase 25 kg more tea for rs . 22500 . what is the reduced price per kg of tea ?
"answer : 180 degree answer : e"
a ) 120 ° , b ) 135 ° , c ) 125 ° , d ) 150 ° , e ) 180 °
e
subtract(multiply(multiply(multiply(const_60, 12), const_2), divide(30, const_60)), multiply(divide(divide(multiply(multiply(const_60, 12), const_2), 12), const_4), add(divide(30, const_60), 12)))
divide(n1,const_60)|multiply(n0,const_60)|add(n0,#0)|multiply(#1,const_2)|divide(#3,n0)|multiply(#0,#3)|divide(#4,const_4)|multiply(#2,#6)|subtract(#5,#7)|
physics
at 12 : 30 , the hour hand and the minute hand of a clock form an angle of
"solution given expression = 896.7 - ( 573.07 + 95.007 ) = 896.7 - 668.077 = 228.623 . answer a"
a ) 228.623 , b ) 224.777 , c ) 233.523 , d ) 414.637 , e ) none of these
a
subtract(896.7, divide(573.07, 95.007))
divide(n1,n2)|subtract(n0,#0)|
general
896.7 – 573.07 – 95.007 = ?
"the dollars earned will be in the same ratio as amount of work done 1 day work of c is 1 / 12 ( or 2 / 24 ) 1 day work of the combined workforce is ( 1 / 6 + 1 / 8 + 1 / 12 ) = 9 / 24 c ' s contribution is 2 / 9 of the combined effort translating effort to $ = 2 / 9 * 2200 = $ 488.9 hence : a"
a ) $ 488.9 , b ) $ 480.9 , c ) $ 588.9 , d ) $ 680.9 , e ) $ 788.9
a
multiply(2200, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8)))))
inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)|
physics
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2200 , what will be c ' s share of the earnings ?
"1 / ( 1 / 36 + 1 / 45 ) = 180 / 9 = 20 hrs answer : a"
a ) 20 hrs , b ) 22 hrs , c ) 23 hrs , d ) 24 hrs , e ) 21 hrs
a
divide(const_1, add(divide(const_1, 36), divide(const_1, 45)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|
physics
two pipes a and b can fill a tank in 36 hours and 45 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ?
"let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 18 the total number of traditionalists is 6 p / 18 = p / 3 . the total population is p + p / 3 = 4 p / 3 p / ( 4 p / 3 ) = 3 / 4 the answer is e ."
a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
e
subtract(1, divide(divide(6, 18), add(divide(6, 18), 1)))
divide(n0,n2)|add(n1,#0)|divide(#0,#1)|subtract(n1,#2)|
general
a certain country is divided into 6 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 18 the total number of progressives in the entire country , what fraction of the country is traditionalist ?
"m = 6.5 s = 1.5 ds = 8 us = 4 as = ( 2 * 8 * 4 ) / 12 = 5.33 answer : a"
a ) 5.33 kmph , b ) 6.00 kmph , c ) 5.00 kmph , d ) 6.00 kmph , e ) 4.00 kmph
a
divide(add(6.5, subtract(6.5, 1.5)), const_2)
subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)|
general
a man whose speed is 6.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ?
"to obtain rs . 10 , investment = rs . 210 . to obtain rs . 1250 , investment = = rs . 26250 . answer : b"
a ) 5363 , b ) 26250 , c ) 28678 , d ) 29002 , e ) 2732
b
multiply(divide(210, 10), 1250)
divide(n2,n1)|multiply(n0,#0)|
gain
in order to obtain an income of rs . 1250 from 10 % stock at rs . 210 , one must make an investment of
"solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 400 x 20.625 ) = rs . 8,250 . answer b"
a ) rs . 15,000 , b ) rs . 8,250 , c ) rs . 15,600 , d ) rs . 16,500 , e ) none
b
multiply(400, multiply(5.5, 3.75))
multiply(n0,n1)|multiply(n2,#0)|
physics
the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 400 per sq . metre .
"solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9632 answer a"
a ) 1.9632 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none
a
multiply(divide(268, const_100), divide(74, const_100))
divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|
general
given that 268 x 74 = 19632 , find the value of 2.68 x . 74 .
xc 1 * ( x - 1 ) c 1 = 56 x ^ 2 - x - 56 = 0 ( x - 8 ) ( x + 7 ) = 0 x = 8 , - 7 - 7 ca n ' t possible . hence 8 should be the answer b
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
b
divide(add(const_1, sqrt(add(multiply(const_4, 56), power(negate(const_1), const_2)))), const_2)
multiply(n0,const_4)|negate(const_1)|power(#1,const_2)|add(#0,#2)|sqrt(#3)|add(#4,const_1)|divide(#5,const_2)
other
a student committee on academic integrity has 56 ways to select a president and vice president from a group of candidates . the same person can not be both president and vice president . how many candidates are there ?
( x - 26,800 ) / 11 - x / 12 = 1,200 x = 480,000 x / 12 = 40,000 answer : c .
a ) $ 29000 , b ) $ 33500 , c ) $ 40000 , d ) $ 41000 , e ) $ 42300
c
add(26800, multiply(const_12, 1200))
multiply(n1,const_12)|add(n0,#0)
general
a businessman earns $ 26800 in december , thus decreasing his average annual ( january to december ) earnings by $ 1200 . his average annual earnings would be source : cmat preparation
let , total work unit = 160 units a can finish in 16 days = 160 unit work i . e . a can finish in 1 days = 10 unit work i . e . b can finish in 1 days = 10 + ( 60 / 100 ) * 10 = 16 unit work days in which b will complete the work alone = 160 / 16 = 10 days answer : option e
a ) 6 , b ) 6.25 , c ) 7 , d ) 7.5 , e ) 10
e
divide(multiply(16, 60), const_100)
multiply(n0,n1)|divide(#0,const_100)
gain
a can complete a certain job in 16 days . b is 60 % more efficient than a . in how many days can b complete the same job ?
"total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 × 600 = 9000 / 100 = 90 number of rotten bananas = 7 % of 400 = 28 therefore , total number of rotten fruits = 90 + 28 = 118 therefore number of fruits in good condition = 1000 - 118 = 882 therefore percentage of fruits in good condition = ( 882 / 1000 × 100 ) % = ( 88200 / 1000 ) % = 88.2 % answer : b"
a ) 92.5 % , b ) 88.2 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 %
b
multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(7, const_100)))), add(600, 400)), const_100)
add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)|
gain
a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 7 % of bananas were rotten . find the percentage of fruits in good condition ?
"let the money sandy took for shopping be x . 0.7 x = 231 x = 330 the answer is d ."
a ) $ 270 , b ) $ 290 , c ) $ 310 , d ) $ 330 , e ) $ 350
d
divide(231, divide(subtract(const_100, 30), const_100))
subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|
gain
sandy had $ 231 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ?
"3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 12 pumps then 48 / 12 = 4 hrs . answer : a"
a ) 4 , b ) 10 , c ) 11 , d ) 12 , e ) 13
a
divide(multiply(multiply(3, 8), 2), 12)
multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|
physics
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 12 pumps work to empty the tank in 1 day ?
"let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 700 gm ) of tea and 1 / 2 kg ( 700 gm ) of coffee ( equal quantities ) = 70 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 70 = > x = 70 thus proved . . . option a ."
a ) 70 , b ) 60 , c ) 80 , d ) 100 , e ) 120
a
divide(70, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1)))
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)|
general
the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 70 / kg . how much did a kg of coffee cost in june ?
let the amounts be x , y , z in ascending order of value . as the interest rate and interest accrued are same for 2 years 6 years and 11 years i . e . 2 x = 6 y = 11 z = k . l . c . m . of 2 , 611 = 66 so x : y : z : = 33000 : 11000 : 6000 the amount deposited for 11 years = 6000 answer : e
a ) 6500 , b ) 2000 , c ) 4500 , d ) 3000 , e ) 6000
e
multiply(multiply(6, const_10), const_100)
multiply(n2,const_10)|multiply(#0,const_100)
general
a sum of rs . 66000 is divided into 3 parts such that the simple interests accrued on them for 6 , two and 11 years respectively may be equal . find the amount deposited for 11 years .
"in 100 consecutive numbers , number of multiples of 5 = 100 / 5 = 20 ( ignore decimals ) in 100 consecutive numbers , number of multiples of 6 = 100 / 6 = 16 number of multiples of 5 * 6 i . e . 30 = 100 / 30 = 3 number of integers from 1 to 100 that are divisible by neither 5 nor by 6 = 100 - ( 29 + 16 - 3 ) { using the concept of sets here ) = 58 answer is c"
a ) 35 , b ) 47 , c ) 58 , d ) 26 , e ) 34
c
subtract(100, subtract(add(divide(100, 5), divide(100, 6)), divide(100, multiply(5, 6))))
divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)|
other
what is the number of integers from 1 to 100 ( inclusive ) that are divisible by neither 5 nor by 6 ?
looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c
a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 %
c
multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100)
add(n0,n1)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|multiply(n0,#2)|subtract(n0,#4)|subtract(#3,#5)|divide(#6,#0)|multiply(#7,const_100)
general
in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what percent z of all the people in smithtown are left - handed women ?
"answer is c . x / x + 36 = 1 / 5 x = 9 16 - 9 = 7"
a ) 3 , b ) 6 , c ) 7 , d ) 7.8 , e ) 9
c
multiply(divide(20, const_100), 20)
divide(n2,const_100)|multiply(n2,#0)|
gain
uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 16 ounces of chocolate . how many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ?
total age of 20 men = 15.6 x 20 = 312 now , total age of 25 men = 364 . total age of five men added later = 364 - 312 = 52 . hence , the total average of five men = 52 / 5 = 10.4 answer : d
a ) 15.5 , b ) 15.4 , c ) 15.25 , d ) 10.4 , e ) 15.6
d
divide(subtract(multiply(add(20, 5), 14.56), multiply(20, 15.6)), 5)
add(n0,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2)
general
the average age of 20 men in the class is 15.6 years . 5 new men join and the new average becomes 14.56 years . what was the average age of 5 new men ?
"let the numbers be x , x + 1 and x + 2 x 2 + ( x + 1 ) 2 + ( x + 2 ) 2 = 2030 3 x 2 + 6 x - 2025 = 0 ( x + 27 ) ( x - 25 ) = 0 x = 25 the middle number is 26 answer b 26"
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29
b
divide(subtract(sqrt(add(multiply(subtract(2030, const_1), const_4), const_1)), const_1), const_2)
subtract(n0,const_1)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|subtract(#3,const_1)|divide(#4,const_2)|
physics
the sum of the squares of three consecutive natural number is 2030 . what is the middle number ?
"d = 130 m + 120 m = 250 m * 1 / 1000 = 0.25 kms rs = 60 + 60 = 120 kmph t = ( 0.25 / 120 ) * 3600 = 7.5 sec answer : e"
a ) 6.9 sec , b ) 7.1 sec , c ) 7.2 sec , d ) 7.4 sec , e ) 7.5 sec
e
divide(add(60, 60), multiply(add(130, 130), const_0_2778))
add(n0,n0)|add(n1,n1)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
in what time will two trains cross each other completely , which are running on the same parallel lines in opposite directions , each train running with a speed of 60 kmph being 130 m and 120 m in length respectively ?
"explanation : increase in the population in 10 years = 2 , 62,500 - 1 , 75,000 = 87500 % ncrease in the population in 10 years = ( 87500 / 175000 ) × 100 = 8750 / 175 = 50 % average % increase of population per year = 50 % / 10 = 5 % answer : option c"
a ) 4 % , b ) 6 % , c ) 5 % , d ) 50 % , e ) none of these
c
add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4)
add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)|
general
the population of a town increased from 1 , 75,000 to 2 , 62,500 in a decade . what is the average percent increase of population per year ?
"age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = ( 225 - 214 ) = 11 years . answer : a"
a ) 11 years , b ) 17 years , c ) 67 years , d ) 14 years , e ) 12 years
a
subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ?
458 - 435 = 23 . answer is c .
a ) 9 , b ) 17 , c ) 23 , d ) 45 , e ) 12
c
subtract(458, 435)
subtract(n1,n0)
physics
rob also compared the empire state building and the petronas towers . what is the height difference between the two if the empire state building is 435 m tall and the petronas towers is 458 m tall ?
"b 5 1 / 4 â € “ 1 / 20 = 1 / 5 = > 5"
a ) 11 , b ) 5 , c ) 6 , d ) 8 , e ) 25
b
add(inverse(subtract(divide(const_1, 4), divide(const_1, 20))), divide(const_2, add(const_2, const_3)))
add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)|
physics
a and b together can do a work in 4 days . if a alone can do it in 20 days . in how many days can b alone do it ?
let x be the value of the item . 0.07 * ( x - 1000 ) = 109.90 x = 2570 the answer is d .
a ) $ 1940 , b ) $ 2150 , c ) $ 2360 , d ) $ 2570 , e ) $ 2780
d
add(1000, divide(109.9, divide(7, const_100)))
divide(n0,const_100)|divide(n2,#0)|add(n1,#1)
general
when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1000 . if the amount of the import tax that the merchant paid was $ 109.90 , what was the total value of the item ?
"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 ) / 34 = 15 = > x = 210 m . answer : e"
a ) 228 , b ) 240 , c ) 887 , d ) 166 , e ) 210
e
multiply(20, multiply(54, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics
a train passes a station platform in 34 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 ?
diameter = 11 meter . radius = diameter / 2 . = 11 / 2 . = 5.5 meter . area of a circle = ï € r 2 . here , pi ( ï € ) = 3.14 meter , radius ( r ) = 5.5 . area of a circle = 3.14 ã — 5.5 ã — 5.5 . . = 3.14 ã — 30.25 . = 95.07 square meter answer : b
['a ) 113.00 square meter', 'b ) 95.07 square meter', 'c ) 93.08 square meter', 'd ) 93.24 square meter', 'e ) 113.43 square meter']
b
circle_area(divide(11, const_2))
divide(n0,const_2)|circle_area(#0)
physics
find the area , diameter = 11 m .
when positive integer n is divided by 3 , the remainder is 1 i . e . , n = 3 x + 1 values of n can be one of { 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 . . . . . . . . . . . . . . 49 , 52 , 59 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 5 , the remainder is 5 . . i . e . , n = 5 y + 4 values of n can be one of { 4 , 9 , 14 , 19 , . . . } combining both the sets we get n = { 4,19 , 52 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 11 or 11 x in case of n = 4 p = 7 so for min value of p , we take min value of n . d is the answer .
a ) 1 , b ) 2 , c ) 5 , d ) 7 , e ) 20
d
subtract(11, reminder(4, 5))
reminder(n3,n2)|subtract(n4,#0)
general
when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 5 , the remainder is 4 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 11 ?
"6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12400 = 3720 . answer : c"
a ) 3630 , b ) 2881 , c ) 3720 , d ) 9977 , e ) 2212
c
multiply(divide(6300, add(add(6300, 4200), 10500)), 12400)
add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|
gain
a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12400 after a year ?
sol . speed of the first train = [ 100 / 10 ] m / sec = 10 m / sec . speed of the second train = [ 100 / 15 ] m / sec = 6.7 m / sec . relative speed = ( 10 + 6.7 ) = m / sec = 16.7 m / sec . ∴ required time = ( 100 + 100 ) / 16.7 secc = 11.9 sec . answer b
a ) 12 , b ) 11.9 , c ) 16 , d ) 20 , e ) 18
b
divide(add(100, 100), add(divide(100, 15), divide(100, 10)))
add(n2,n2)|divide(n2,n1)|divide(n2,n0)|add(#1,#2)|divide(#0,#3)
physics
two tains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each train be 100 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?
"40 % of ( x - y ) = 20 % of ( x + y ) 40 / 100 ( x - y ) = 20 / 100 ( x + y ) x = 3 y required percentage = y / x * 100 = y / 3 y * 100 = 33.3 % answer is d"
a ) 50.5 % , b ) 44.4 % , c ) 22.2 % , d ) 33.3 % , e ) 25 %
d
multiply(divide(subtract(40, 20), add(40, 20)), const_100)
add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|
general
if 40 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ?
"to reach the winning post a will have to cover a distance of ( 500 - 170 ) m , i . e . , 330 m . while a covers 3 m , b covers 4 m . while a covers 330 m , b covers 4 x 330 / 3 m = 440 m . thus , when a reaches the winning post , b covers 440 m and therefore remains 60 m behind . a wins by 60 m . answer : a"
a ) 60 m , b ) 20 m , c ) 43 m , d ) 20 m , e ) 23 m
a
subtract(500, divide(multiply(subtract(500, 170), 4), 3))
subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)|
physics
in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 170 m . then , a wins by :
p ( club ) = 1 / 4 p ( red king ) = 1 / 26 p ( club then a red king ) = 1 / 4 * 1 / 26 = 1 / 104 the answer is e .
a ) 1 / 13 , b ) 1 / 15 , c ) 1 / 26 , d ) 1 / 52 , e ) 1 / 104
e
multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52))
divide(const_2,const_52)|multiply(const_3,const_4)|add(#1,const_1)|divide(#2,const_52)|multiply(#3,#0)
probability
from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a red king ?
the mini - cubes with 2 painted sides are all on the edge of the cube , in the ` ` middle ' ' of the edge . there are 4 in front , 4 in back and 4 more on the ` ` strip ' ' that runs around the left / top / right / bottom of the cube . 4 + 4 + 4 = 12 . answer a
['a ) 12', 'b ) 8', 'c ) 6', 'd ) 10', 'e ) 16']
a
multiply(const_4, power(27, divide(const_1, const_3)))
divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)
geometry
a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many cubes are painted on only 2 faces ?
"total fish = x percentage of second catch = ( 2 / 40 ) * 100 = 5 % so , x * 5 % = 50 x = 1000 ans . a"
a ) 1000 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 10,000
a
divide(40, divide(2, 40))
divide(n2,n1)|divide(n0,#0)|
gain
in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 40 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 ?
"explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 3 − 1 / 6 ) = ( 6 − 3 ) / 18 = 1 / 6 so son will do whole work in 6 days answer : b"
a ) 7 days , b ) 6 days , c ) 5 days , d ) 4 days , e ) none of these
b
divide(multiply(6, 3), subtract(6, 3))
multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
physics
a man can do a piece of work in 6 days , but with the help of his son he can do it in 3 days . in what time can the son do it alone ?
"1 - 9 = > 1 * 9 digits 10 - 99 = > 2 * 90 = 180 ( numbers between 10 - 99 is 90 where each has 2 digits ) 100 - 999 = > 3 * 900 = 2700 1000 - 9999 = > 4 * 9000 = 36,000 10000 - 59999 = > 5 * 50,000 = 250,000 the answer is 288,889 the answer is c ."
a ) 248,889 , b ) 268,889 , c ) 288,889 , d ) 308,889 , e ) 328,889
c
add(add(add(subtract(const_10, 1), multiply(subtract(const_100, const_10), const_2)), multiply(subtract(const_100, const_10), const_3)), multiply(subtract(const_100, const_10), const_4))
subtract(const_100,const_10)|subtract(const_10,n0)|multiply(#0,const_2)|multiply(#0,const_3)|multiply(#0,const_4)|add(#2,#1)|add(#5,#3)|add(#6,#4)|
general
meena wrote all the numbers from 1 to 59,999 inclusive . how many digits did she write in total ?
"3 < x < 6 < y < 9 ; 3 < x y < 9 3 + y < x + 9 y - x < 6 . positive integer difference is 5 ( for example y = 8.5 and x = 3.5 ) answer : c ."
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
c
subtract(subtract(9, 3), const_1)
subtract(n2,n0)|subtract(#0,const_1)|
general
if 3 < x < 6 < y < 9 , then what is the greatest possible positive integer difference of x and y ?
"imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 20 % of 350 , comes out to be 70 . . . . 150 % of 10 is 15 . . . . ( 70 * 100 ) / 15 = 466.66 . . . . ans : b"
a ) 0.35 , b ) 466 , c ) 35 , d ) 350 , e ) 3500
b
multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 20), 150), const_100)
multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100)|
gain
if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 150 % of j ?
"explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 22 ) = rs . 176 . profit = 176 - 150 = 26 so , profit % = ( 26 / 150 ) * 100 = 17 % option e"
a ) 12 % , b ) 13 % , c ) 14 % , d ) 15 % , e ) 17 %
e
divide(multiply(subtract(multiply(22, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20)))
add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3)|
gain
a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 22 per kg , then his gain percent is
"explanation : let the cost price of one article = rs . 1 cp of x articles = rs . x cp of 20 articles = 20 selling price of x articles = 20 profit = 25 % [ given ] ⇒ ( sp − cp / cp ) = 25 / 100 = 1 / 4 ⇒ ( 20 − x ) / x = 1 / 4 ⇒ 80 − 4 x = x ⇒ 5 x = 80 option d ⇒ x = 805 = 16"
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17
d
divide(multiply(20, const_4), add(const_4, const_1))
add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|
gain
the cost price of 20 articles is the same as the selling price of x articles . if the profit is 25 % , find out the value of x
"19 / 36 m / s = 19 / 36 * 18 / 5 = 19 / 10 = 1.9 kmph . answer : e"
a ) 1.7 , b ) 1.5 , c ) 1.3 , d ) 1.1 , e ) 1.9
e
multiply(const_3_6, divide(19, 36))
divide(n0,n1)|multiply(#0,const_3_6)|
physics
convert the 19 / 36 m / s into kilometers per hour ?
"percentage error in calculated area = ( 5 + 5 + ( 5 ã — 5 ) / 100 ) % = 10.25 % answer : d"
a ) 4.05 % , b ) 4.02 % , c ) 4 % , d ) 10.28 % , e ) 2 %
d
divide(multiply(subtract(square_area(add(const_100, 5)), square_area(const_100)), const_100), square_area(const_100))
add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|
gain
an error 5 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ?
"from 9 : 00 in the morning to 6 : 00 in the afternoon , inclusive there are 9 * 10 = 90 five - minute intervals , thus total of 54 * 30 tickets were sold . say x student and 2 x regular tickets were sold , then x + 2 x = 90 * 30 - - > x = 30 * 30 and 2 x = 2 * ( 30 * 30 ) = 30 * 60 . therefore , the total revenue from ticket sales that day was 30 * 30 * 6 + 30 * 60 * 10 = $ 23,400 . answer : a ."
a ) $ 23400 , b ) $ 25920 , c ) $ 28080 , d ) $ 28500 , e ) $ 29160
a
add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, 6), 9), const_1), const_12), 30), add(const_3.0, const_1)), 2), 10), multiply(divide(multiply(multiply(add(subtract(add(const_12, 6), 9), const_1), const_12), 30), add(2, const_1)), 6))
add(const_12,n4)|add(const_3.0,const_1)|subtract(#0,n2)|add(#2,const_1)|multiply(#3,const_12)|multiply(n0,#4)|divide(#5,#1)|multiply(n8,#6)|multiply(n7,#6)|multiply(n6,#7)|add(#9,#8)|
general
for a certain art exhibit , a museum sold admission tickets to a group of 30 people every 6 minutes from 9 : 00 in the morning to 6 : 00 in the afternoon , inclusive . the price of a regular admission ticket was $ 10 and the price of a student ticket was $ 6 . if on one day 2 times as many regular admission tickets were sold as student tickets , what was the total revenue from ticket sales that day ?
"explanation : a runs 1000 meters while b runs 900 meters and c runs 830 meters . therefore , b runs 900 meters while c runs 830 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 830 ) / 900 = 922.22 meters thus , b can give c ( 1000 - 922.22 ) = 77.77 meters start answer : c"
a ) 11.77 meters , b ) 55.77 meters , c ) 77.77 meters , d ) 113.77 meters , e ) none of these
c
subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 170)), subtract(multiply(const_100, const_10), 100)))
multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)|
physics
a can give b 100 meters start and c 170 meters start in a kilometer race . how much start can b give c in a kilometer race ?
# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 7 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 7 / 5 = 7 - - > total time for both of them would be 10 + 7 = 17 - - > deluxe / total = 7 / 17 . b
a ) 5 / 17 , b ) 7 / 17 , c ) 4 / 17 , d ) 3 / 17 , e ) 5
b
divide(multiply(5, divide(7, 5)), add(multiply(multiply(2, 5), const_1), multiply(5, divide(7, 5))))
divide(n2,n3)|multiply(n0,n3)|multiply(n3,#0)|multiply(#1,const_1)|add(#3,#2)|divide(#2,#4)
general
company t produces two kinds of stereos : basic and deluxe . of the stereos produced by company t last month , 2 / 3 were basic and the rest were deluxe . if it takes 7 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ?
"volume of the block = 10 * 20 * 30 = 6000 cm ^ 3 side of the largest cube = h . c . f of 10 , 20,30 = 10 cm volume of the cube = 10 * 10 * 10 = 1000 cm ^ 3 number of cubes = 6000 / 1000 = 6 answer is a"
a ) 6 , b ) 10 , c ) 15 , d ) 40 , e ) 22
a
divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(30, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(30, const_10))), const_10))
multiply(const_1000,const_3)|multiply(n2,const_10)|multiply(const_2,const_3)|rectangle_area(const_1000,const_360)|add(#0,#1)|divide(#3,const_10)|multiply(#2,const_1000)|subtract(#5,#6)|add(#4,#7)|divide(#8,const_10)|divide(#8,#9)|
geometry
a rectangular block 10 cm by 20 cm by 30 cm is cut into an exact number of equal cubes . find the least possible number of cubes ?
oa is definitely wrong . the answer should be e .
a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75
e
add(multiply(0.25, subtract(5, 1)), 0.25)
subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)|
general
at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 11.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ?
"sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 45 % = 45 % of 640 = rs . 288 sp = cp - loss = 640 - 288 = rs . 352 answer : b"
a ) s . 429 , b ) s . 352 , c ) s . 429 , d ) s . 128 , e ) s . 419
b
subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 45), const_100))
add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 45 % ?
"i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 5 ) = $ 200 total revenue : 1000 + 200 = $ 1200 the answer is b ."
a ) $ 1220 , b ) $ 1200 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880
b
add(multiply(add(10, 5), 30), multiply(subtract(100, 30), 10))
add(n0,n1)|subtract(n2,n3)|multiply(n3,#0)|multiply(n0,#1)|add(#2,#3)|
physics
at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 5 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ?
"70 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 126 = > x = 14 28 : 14 : 84 84 - 14 = 70 answer d"
a ) 18 , b ) 36 , c ) 72 , d ) 70 , e ) 108
d
subtract(divide(126, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(126, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2)))
divide(n1,n2)|multiply(n2,const_2)|add(n1,#0)|divide(n1,#1)|add(#2,#3)|divide(n0,#4)|divide(#5,#1)|subtract(#5,#6)|
general
pat , kate , and mark charged a total of 126 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much time as mark , how many more hours did mark charge to the project than kate ?
"the triangle with sides 52 cm , 48 cm and 20 cm is right angled , where the hypotenuse is 52 cm . area of the triangle = 1 / 2 * 48 * 20 = 480 cm 2 answer : a"
a ) 480 cm 2 , b ) 580 cm 2 , c ) 380 cm 2 , d ) 180 cm 2 , e ) 280 cm 2
a
multiply(divide(48, const_2), 20)
divide(n1,const_2)|multiply(n2,#0)|
geometry
calculate the area of a triangle , if the sides of are 52 cm , 48 cm and 20 cm , what is its area ?
"there are 7 ! ways to make codes starting with 12 . there are 7 ! ways to make codes starting with 24 . there are 7 ! ways to make codes starting with 36 . there are 7 ! ways to make codes starting with 48 . the number of codes is 4 * 7 ! = 20,160 . the answer is d ."
a ) 14,720 , b ) 16,240 , c ) 18,320 , d ) 20,160 , e ) 22,480
d
multiply(multiply(2, 3), 2)
multiply(n3,n4)|multiply(n3,#0)|
general
employees of a certain company are each to receive a unique 9 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , and 8 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ?
let the persons be a , b , c . hours worked : a = 2 * 144 / 12 = 24 hours b = 4 * 144 / 12 = 48 hours c = 6 * 144 / 12 = 72 hours c is the hardest worker and a worked for the least number of hours . so the difference is 72 - 24 = 48 hours . answer : c
a ) 47 hours , b ) 45 hours , c ) 48 hours , d ) 49 hours , e ) 50 hours
c
subtract(multiply(divide(144, add(add(2, 4), 6)), 6), multiply(divide(144, add(add(2, 4), 6)), 2))
add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|multiply(n0,#2)|subtract(#3,#4)
physics
the amount of time that three people worked on a special project was in the ratio of 2 to 4 to 6 . if the project took 144 hours , how many more hours did the hardest working person work than the person who worked the least ?
"take the multiples of 7 and add 5 0 x 7 + 5 = 5 . . . . 6 x 7 + 5 = 47 there are 7 numbers answer b"
a ) 3 , b ) 7 , c ) 4 , d ) 5 , e ) 6
b
divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
general
how many positive integers less than 50 have a reminder 5 when divided by 7 ?
"explanation : cash price = $ 24 000 deposit = 10 % ã — $ 24 000 = $ 2400 loan amount = $ 24000 â ˆ ’ $ 2400 number of payments = 60 = $ 21600 i = p * r * t / 100 i = 12960 total amount = 21600 + 12960 = $ 34560 regular payment = total amount / number of payments = 576 answer : d"
a ) $ 503 , b ) $ 504 , c ) $ 555 , d ) $ 576 , e ) $ 587
d
add(divide(multiply(multiply(24, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(24, const_1000), subtract(const_1, divide(10, const_100)))))
divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)|
gain
a car is purchased on hire - purchase . the cash price is $ 24 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ?
"explanation : speed of the truck = distance / time = 550 / 1 = 550 meters / minute speed of the train = distance / time = 33 / 45 km / minute = 33000 / 45 meters / minut speed of the truck / speed of the train = 550 / ( 33000 / 45 ) = ( 550 × 45 ) / 33000 = ( 55 × 45 ) / 3300 = ( 11 × 45 ) / 660 = ( 11 × 9 ) / 132 = 9 / 12 = 34 hence , speed of the truck : speed of the train = 3 : 4 answer : option d"
a ) 3 : 7 , b ) 4 : 7 , c ) 1 : 4 , d ) 3 : 4 , e ) 2 : 5
d
divide(550, multiply(divide(33, 45), const_1000))
divide(n2,n3)|multiply(#0,const_1000)|divide(n0,#1)|
physics
a truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes . what is the ratio of their speed ?
"let the shares of a , b , c , d are 6 x , 4 x , 8 x , 5 x 8 x - 5 x = 3000 3 x = 3000 , x = 1000 b ' s share = 4 x = $ 4000 answer is d"
a ) $ 2000 , b ) $ 6000 , c ) $ 1000 , d ) $ 4000 , e ) $ 5000
d
divide(multiply(divide(multiply(add(3000, 3000), 5), 8), 4), 5)
add(n4,n4)|multiply(n3,#0)|divide(#1,n2)|multiply(n1,#2)|divide(#3,n3)|
general
a sum of money is distributed among a , b , c , d in the proportion of 6 : 4 : 8 : 5 . if c gets $ 3000 more than d , what is the b ' s share ?
"explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 380 ) = 2 + 2 ( extra ) = > 2 x - 760 = x = > x = 760 . answer : c"
a ) 237 , b ) 287 , c ) 760 , d ) 287 , e ) 720
c
multiply(380, const_2)
multiply(n0,const_2)|
general
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 380 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ?
"the percent of the population who are employed females is 70 - 42 = 28 % the percent of employed people who are female is 28 % / 70 % = 40 % . the answer is d ."
a ) 25 % , b ) 30 % , c ) 35 % , d ) 40 % , e ) 45 %
d
multiply(divide(subtract(70, 42), 70), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
in town p , 70 percent of the population are employed , and 42 percent of the population are employed males . what percent of the employed people in town p are females ?
"initial no of students + 3 * ( 1 + no of possible 3 minute intervals between 15 : 03 and 15 : 44 ) - 8 * ( 1 + no of possible 10 minute intervals between 15 : 10 and 15 : 44 ) 20 + 3 * 14 - 8 * 4 = 25 c"
a ) 7 , b ) 14 , c ) 25 , d ) 27 , e ) 30
c
add(subtract(add(multiply(floor(divide(44, 03)), 03), 20), multiply(floor(divide(44, 9)), 9)), 03)
divide(n10,n4)|divide(n10,n8)|floor(#0)|floor(#1)|multiply(n4,#2)|multiply(n8,#3)|add(n2,#4)|subtract(#6,#5)|add(n4,#7)|
physics
at 15 : 00 there were 20 students in the computer lab . at 15 : 03 and every three minutes after that , 3 students entered the lab . if at 15 : 10 and every ten minutes after that 9 students left the lab , how many students were in the computer lab at 15 : 44 ?
"option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than 0.6 x 14 + 13 ( 21.4 ) . . . it ' s pretty close so let ' s see if one more day will make it worth it . . . if we raise the number of days to 15 , the sticker option looks like a better deal . . . 1.5 x 15 ( 22.5 ) vs 0.6 x 15 + 13 ( 22 ) . answer : c"
a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29
c
add(multiply(divide(multiply(divide(13.00, multiply(subtract(0.65, 0.30), const_2)), const_2), const_10), const_2), multiply(divide(13.00, multiply(subtract(0.65, 0.30), const_2)), const_2))
subtract(n0,n2)|multiply(#0,const_2)|divide(n1,#1)|multiply(#2,const_2)|divide(#3,const_10)|multiply(#4,const_2)|add(#5,#3)|
general
the toll for crossing a certain bridge is $ 0.65 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 13.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day , what is the least value of x for which this driver can save money by using the sticker ?
"the volume of water in the tank is h * l * b = 75 cubic feet . since h = 3 , then l * b = 25 and l = b = 5 . since the tank is cubical , the capacity of the tank is 5 * 5 * 5 = 125 . the ratio of the water in the tank to the capacity is 75 / 125 = 3 / 5 the answer is e ."
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 2 / 5 , e ) 3 / 5
e
divide(3, divide(75, const_10))
divide(n1,const_10)|divide(n0,#0)|
physics
a cubical tank is filled with water to a level of 3 feet . if the water in the tank occupies 75 cubic feet , to what fraction of its capacity is the tank filled with water ?
"135 - 74 + 1 = 62 ' b ' is the answer"
a ) 53 , b ) 62 , c ) 51 , d ) 50 , e ) 49
b
add(subtract(135, 74), const_1)
subtract(n1,n0)|add(#0,const_1)|
general
andy solves problems 74 to 135 inclusive in a math exercise . how many problems does he solve ?
"answer is 9 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 9009 / 1001 = 9 answer d"
a ) 0.009 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90
d
multiply(divide(9.009, 1.001), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
9.009 / 1.001
"formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 25 % = 125 % 125 % - - - - - - - > 520 ( 120 ã — 4.16 = 520 ) 100 % - - - - - - - > 416 ( 100 ã — 4.16 = 416 ) option ' e '"
a ) 216 , b ) 316 , c ) 616 , d ) 516 , e ) 416
e
divide(520, add(const_1, divide(25, const_100)))
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
gain
a number increased by 25 % gives 520 . the number is ?
"x / 14 + x / 30 + x / 40 = 1 x = 7.7 days answer : b"
a ) 7.6 days , b ) 7.7 days , c ) 6.7 days , d ) 5.7 days , e ) 8.7 days
b
add(subtract(subtract(14, subtract(40, 30)), 4), const_1)
subtract(n2,n1)|subtract(n0,#0)|subtract(#1,n3)|add(#2,const_1)|
physics
a , b and c can do a piece of work in 14 , 30 and 40 days respectively . they start the work together but c leaves 4 days before the completion of the work . in how many days is the work done ?
"t 1 = 240 / 60 = 4 hours t 2 = 120 / 40 = 3 hours t = t 1 + t 2 = 7 hours avg speed = total distance / t = 360 / 7 = 51 mph = b"
a ) 42 , b ) 51 , c ) 50 , d ) 54 , e ) 56
b
divide(add(240, 120), add(divide(240, 60), divide(120, 40)))
add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)|
physics
joe drives 240 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ?
"the fastest way in an ap is to find the average and multiply with total integers . . between 38 and 127 , the smallest multiple of 4 is 40 and largest = 124 . . average = ( 40 + 124 ) / 2 = 164 / 2 = 82 . . total numbers = ( 124 - 40 ) / 4 + 1 = = 84 / 4 + 1 = 27 + 1 = 22 . . sum = 82 * 22 = 1804 ans a"
a ) 1804 , b ) 1816 , c ) 1824 , d ) 1828 , e ) 1832
a
multiply(divide(add(subtract(127, const_3), add(38, const_2)), const_2), add(divide(subtract(subtract(127, const_3), add(38, const_2)), 4), const_1))
add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)|
general
what is the sum of the multiples of 4 between 38 and 127 inclusive ?
"ans is d given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 10 ) + ( x / 18 ) ) = 14.5 ( formula avg speed = total distance / total time )"
a ) 13 , b ) 13.5 , c ) 14 , d ) 14.5 , e ) 15
d
divide(add(multiply(18, const_10), divide(multiply(18, const_10), const_2)), add(divide(multiply(18, const_10), 10), divide(divide(multiply(18, const_10), const_2), 18)))
multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)|
general
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 10 miles per gallon , and from town b to town c , the car averaged 18 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ?
"speed of train relative to man = 30 + 6 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . time taken to pass the men = 110 / 10 = 11 sec . answer : d"
a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 11 sec , e ) 2 sec
d
divide(110, multiply(add(30, 6), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
a train 110 m long is running with a speed of 30 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ?
here one boy is excluded and final average of the group decreases . ∴ change in average is ( – ) ve = – 0.1 kg . using the formula sum of the quantities excluded = ( changein no . ofquantities × origina laverage ) + ( changeinaverage × final no . ofquantities ) ⇒ weight of the boy who left = ( 1 × 45 ) – ( – 0.1 × 49 ) = 49.9 kg answer c
a ) 40.9 kg , b ) 42.9 kg , c ) 49.9 kg , d ) 39.9 kg , e ) none of these
c
add(45, divide(multiply(subtract(50, const_1), 100), const_1000))
subtract(n0,const_1)|multiply(n2,#0)|divide(#1,const_1000)|add(n1,#2)
general
there are 50 boys in a class . their average weight is 45 kg . when one boy leaves the class , the average reduces by 100 g . find the weight of the boy who left the class .
"1 h - - - - - 5 ? - - - - - - 60 12 h rs = 20 + 21 = 41 t = 12 d = 41 * 12 = 492 answer : b"
a ) 288 , b ) 492 , c ) 877 , d ) 278 , e ) 178
b
add(multiply(divide(60, subtract(21, 20)), 20), multiply(divide(60, subtract(21, 20)), 21))
subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
physics
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 20 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ?
from here , it might be easier to go up in bounds of 60 , so we know that 61 - 120 gives 10 more numbers . 121 - 180 and 181 - 240 as well . this brings us up to 240 with 40 numbers . a cursory glance at the answer choices should confirm that it must be 42 , as all the other choices are very far away . answer choice a is correct here .
a ) 40 , b ) 31 , c ) 42 , d ) 53 , e ) 64
a
divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
general
how many positive integers less than 240 are multiple of 4 but not multiples of 6 ?
"w / x = 1 / 3 = > x = 3 w and w / y = 4 / 15 = > y = 15 / 4 w ( x + y ) / y = ( 3 w + 15 / 4 w ) / ( 15 / 4 w ) = ( 27 / 4 w ) / ( 15 / 4 w ) = 9 / 5 correct option : e"
a ) 4 / 5 , b ) 6 / 5 , c ) 7 / 5 , d ) 8 / 5 , e ) 9 / 5
e
add(divide(divide(4, 1), divide(15, 3)), const_1)
divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)|
general
if w / x = 1 / 3 and w / y = 4 / 15 , then ( x + y ) / y =
"price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 22 tickets = 20 * 22 = 440 revenue generated from sales of 52 tickets = 120 + 340 + 440 = 900 $ answer d"
a ) $ 600 , b ) $ 740 , c ) $ 850 , d ) $ 900 , e ) $ 1,140
d
multiply(add(add(subtract(subtract(52, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20)
divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n0)|subtract(const_1,#0)|subtract(#2,n1)|subtract(const_1,#1)|multiply(n1,#3)|multiply(n0,#5)|add(#6,#4)|add(#8,#7)|multiply(n0,#9)|
gain
tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 52 people bought tickets to the concert , what was the total revenue from ticket sales ?
"lets say x ounces of p is mixed with q . = > 64 - x ounces of q is present in the mixture ( as the total = 64 ounces ) given total almond weight = 14 ounces ( 20 x / 100 ) + ( 25 / 100 ) ( 64 - x ) = 14 = > x = 40 = > 64 - 40 = 14 ounces of q is present in the mixture . answer is a ."
a ) 14 , b ) 20 , c ) 32 , d ) 44 , e ) 48
a
divide(subtract(14, multiply(divide(20, const_100), 64)), subtract(divide(25, const_100), divide(20, const_100)))
divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n3,#2)|divide(#4,#3)|
general
a bowl of nuts is prepared for a party . brand p mixed nuts are 20 % almonds and brand q ' s deluxe nuts are 25 % almonds . if a bowl contains a total of 64 ounces of nuts , representing a mixture of both brands , and 14 ounces of the mixture are almonds , how many ounces of brand q ' s deluxe mixed nuts are used ?
"explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 3 / 60 ) km = 1 / 20 km = 50 m . . ' . distance between the criminal and policeman = ( 265 - 50 ) m = 215 m . answer : e"
a ) 100 m , b ) 120 m , c ) 130 m , d ) 150 m , e ) none of these
e
subtract(265, multiply(divide(3, const_60), const_1000))
divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)|
physics
a policeman noticed a criminal from a distance of 265 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 3 minutes ?
6 flavours * 6 choices = 6 c 1 * 6 c 1 = 6 * 6 = 36 = d
a ) 4 , b ) 8 , c ) 12 , d ) 36 , e ) 32
d
multiply(6, 6)
multiply(n0,n0)
general
a university cafeteria offers 6 flavors of pizza - pork , gobi - manjurian , pepperoni , chicken , hawaiian and vegetarian . if a customer has an option ( but not the obligation ) to add extra cheese , mushrooms or both to any kind of pizza , how many different pizza varieties are available ?
"let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 117 9 x = 108 x = 12 largest number = 3 x + 6 = 42 answer : d"
a ) 45 , b ) 48 , c ) 51 , d ) 42 , e ) 54
d
add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2)))
subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)|
general
the sum of three consecutive multiples of 3 is 117 . what is the largest number ?
one short cut to solve the problem is c : p = 1 : 3 c increased to 5 = > 1 : 3 = 5 : x = > x = 15 = > p increased by 12 b is the answer
a ) 2 , b ) 12 , c ) 5 , d ) 10 , e ) 15
b
subtract(multiply(4, 4), const_4)
multiply(n0,n0)|subtract(#0,const_4)
other
in a zoo , the ratio of the number of cheetahs to the number 4 then what is the increase in the number of pandas ?
"d = 72 * 5 / 18 = 50 = 1000 â € “ 250 = 750 m answer : d"
a ) 150 m , b ) 200 m , c ) 250 m , d ) 750 m , e ) 300 m
d
subtract(multiply(50, multiply(72, const_0_2778)), 250)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics
a train 250 m long running at 72 kmph crosses a platform in 50 sec . what is the length of the platform ?
"3 multiples are . . . 6,9 , 12,15 , 18,21 , 24,27 , 30,33 , 36,39 , 42,45 , 48,51 , 54,57 , . . . , the answer is = 18 answer is c"
a ) 16 , b ) 22 , c ) 18 , d ) 11 , e ) 9
c
add(divide(subtract(59, 5), 3), const_1)
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
general
how many multiples of 3 are there between 5 and 59 , 5 and 59 inclusive ?
"given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 12 ) = 36 / 21 = 12 / 7 . answer : a ."
a ) 12 / 7 , b ) 4 / 1 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 2
a
divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 12))
add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)|
general
in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 12 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture .
"number of passengers using kennedy airport = 32 / 3 = ~ 10.67 passengers using miami airport = 10.67 / 2 = ~ 5.34 passengers using logan airport = 5.34 / 4 = ~ 1.33 so d"
a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 1.33 , e ) 1.6
d
divide(divide(32.3, 3), multiply(4, 2))
divide(n3,n2)|multiply(n5,n6)|divide(#0,#1)|
general
in 1979 approximately 1 / 3 of the 32.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 2 the number that used kennedy airport and 4 times the number that used logan airport , approximately how many millions of these passengers used logan airport that year ?
"let the greater and the smaller number be g and s respectively . gs = 2496 g + s exceeds g - s by 64 i . e . , g + s - ( g - s ) = 64 i . e . , 2 s = 64 = > s = 32 . g = 2496 / s = 78 . answer : d"
a ) a ) 96 , b ) b ) 108 , c ) c ) 110 , d ) d ) 78 , e ) of these
d
divide(2496, multiply(power(const_2, const_4), const_2))
power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)|
general
what is the greater of the two numbers whose product is 2496 , given that the sum of the two numbers exceeds their difference by 64 ?
"a + b + c = 6490 4 a = 6 b = 2 c = x a : b : c = 1 / 4 : 1 / 6 : 1 / 2 = 3 : 2 : 6 3 / 11 * 6490 = rs 1770 answer : e"
a ) s 6490 , b ) s 1880 , c ) s 1660 , d ) s 1550 , e ) s 1770
e
multiply(4, divide(6490, add(add(4, 2), const_3)))
add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)|
general
rs . 6490 is divided so that 4 times the first share , six times the 2 nd share and twice the third share amount to the same . what is the value of the first share ?
of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 120 geese survived the first year : 1 / 5 * ( total ) = 125 - - > ( total ) = 625 . answer : d .
a ) 280 , b ) 400 , c ) 540 , d ) 625 , e ) 840
d
divide(divide(divide(125, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(const_2, const_3))
divide(n1,n5)|divide(n1,n3)|divide(const_2,const_3)|subtract(const_1,#0)|divide(n6,#3)|divide(#4,#1)|divide(#5,#2)
general
of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 125 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ?
explanation : salary of the manager = ( 56 * 8800 - 55 * 8500 ) = 25300 answer : d
a ) 10000 , b ) 12000 , c ) 23000 , d ) 25300 , e ) 45000
d
subtract(multiply(add(55, const_1), 8800), multiply(55, 8500))
add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)
general
the average salary per month of 55 employees in a company is rs 8500 . if the managers salary is added , the average salary increases to rs 8800 , what is the salary of the manager ?
"let x be the number of roses . then the number of birches is 24 − x , and the number of boys is 3 × ( 24 − x ) . if each girl planted 3 roses , there are x 3 girls in the class . we know that there are 24 students in the class . therefore x 3 + 3 ( 24 − x ) = 24 x + 9 ( 24 − x ) = 3 ⋅ 24 x + 216 − 9 x = 72 216 − 72 = 8 x 1448 = x 1 x = 18 so , students planted 18 roses and 24 - x = 24 - 18 = 6 birches . correct answer is d ) 6"
a ) 2 , b ) 5 , c ) 8 , d ) 6 , e ) 4
d
divide(subtract(multiply(3, 24), 24), subtract(multiply(3, 3), 1))
multiply(n0,n1)|multiply(n1,n1)|subtract(#0,n0)|subtract(#1,n2)|divide(#2,#3)|
physics
there are 24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted 3 roses , every three boys planted 1 birch . by the end of the day they planted 2424 plants . how many birches were planted ?
"factors of 12 - 1 , 2 , 3 , 4 , 6 , and 12 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 3 answer c"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
c
divide(16, 10)
divide(n1,n0)|
other
how many of the positive factors of 10 , 16 and how many common factors are there in numbers ?
here n ( s ) = ( 6 * 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } p ( e ) = n ( e ) / n ( s ) = 15 / 36 = 5 / 12 option c
a ) 5 / 7 , b ) 4 / 7 , c ) 5 / 12 , d ) 4 / 7 , e ) 1 / 6
c
divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2)))
add(n0,const_4)|add(const_2,const_4)|add(#0,const_4)|multiply(#1,#1)|divide(#2,#3)
general
in a simultaneous throw of pair of dice . find the probability of getting the total more than 7
"largest 4 digit number is 999999 after doing 999999 ÷ 96 we get remainder 55 hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944 answer : d"
a ) 999991 , b ) 999965 , c ) 999912 , d ) 999936 , e ) 999930
d
multiply(add(const_100, const_2), 99)
add(const_100,const_2)|multiply(n1,#0)|
general