Problem
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628
| Rationale
stringlengths 1
2.74k
| options
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| correct
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848
| linear_formula
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|---|---|---|---|---|---|---|
on dividing 123 by a number , the quotient is 7 and the remainder is 4 . find the divisor .
|
"d = ( d - r ) / q = ( 123 - 4 ) / 7 = 119 / 7 = 17 c"
|
a ) 1 / 2 , b ) 130 cm , c ) 17 , d ) 8 and 20 , e ) 400
|
c
|
floor(divide(123, 7))
|
divide(n0,n1)|floor(#0)|
|
general
|
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 ?
|
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 . 90 , b ) 22.22 meters , c ) 66.67 % , d ) 762 , e ) 350 m
|
a
|
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
|
at 12 : 30 , the hour hand and the minute hand of a clock form an angle of
|
"answer : 180 degree answer : e"
|
a ) 9 , b ) $ 28.44 , c ) 90 , d ) 180 ° , e ) 175.5 cm
|
d
|
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
|
896.7 – 573.07 – 95.007 = ?
|
"solution given expression = 896.7 - ( 573.07 + 95.007 ) = 896.7 - 668.077 = 228.623 . answer a"
|
a ) 5 , b ) 50 % , c ) 10 min , d ) 34 , e ) 228.623
|
e
|
subtract(896.7, divide(573.07, 95.007))
|
divide(n1,n2)|subtract(n0,#0)|
|
general
|
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 ?
|
"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 ) 6 , c ) 17 , d ) 7 , e ) 1
|
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
|
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 ?
|
"1 / ( 1 / 36 + 1 / 45 ) = 180 / 9 = 20 hrs answer : a"
|
a ) 20 hrs , b ) 9 : 16 , c ) 11 % , d ) 1.6 , e ) 725117481
|
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
|
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 ?
|
"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 ) 70400 yards , b ) 67 % , c ) 40 , d ) 3 / 4 , e ) 24 days
|
d
|
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 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 ?
|
"m = 6.5 s = 1.5 ds = 8 us = 4 as = ( 2 * 8 * 4 ) / 12 = 5.33 answer : a"
|
a ) 10 v 2', ' , b ) 5 % , c ) 5.33 kmph , d ) 75 % decrease , e ) 4000
|
c
|
divide(add(6.5, subtract(6.5, 1.5)), const_2)
|
subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)|
|
general
|
in order to obtain an income of rs . 1250 from 10 % stock at rs . 210 , one must make an investment of
|
"to obtain rs . 10 , investment = rs . 210 . to obtain rs . 1250 , investment = = rs . 26250 . answer : b"
|
a ) 1 , b ) 84 % , c ) 12 cm', ' , d ) 26250 , e ) 12
|
d
|
multiply(divide(210, 10), 1250)
|
divide(n2,n1)|multiply(n0,#0)|
|
gain
|
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 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 . 8,250 , b ) 50 km / hr , c ) 1 : 1 , d ) 13 , e ) 1.5 days
|
a
|
multiply(400, multiply(5.5, 3.75))
|
multiply(n0,n1)|multiply(n2,#0)|
|
physics
|
given that 268 x 74 = 19632 , find the value of 2.68 x . 74 .
|
"solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9632 answer a"
|
a ) 64 , b ) 30 minutes , c ) 11 / 30 , d ) 1.9632 , e ) 361
|
d
|
multiply(divide(268, const_100), divide(74, const_100))
|
divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|
|
general
|
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 ?
|
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 ) 67 % , b ) 70 % , c ) rs . 937.5 , d ) 8 , e ) 3
|
d
|
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 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
|
( x - 26,800 ) / 11 - x / 12 = 1,200 x = 480,000 x / 12 = 40,000 answer : c .
|
a ) 8.33 km , b ) − 2 % , c ) 46 , d ) $ 40000 , e ) 90.09 meters
|
d
|
add(26800, multiply(const_12, 1200))
|
multiply(n1,const_12)|add(n0,#0)
|
general
|
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 ?
|
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 ) 78 + ( 4 / , b ) 40 , c ) 0.05 , d ) 10 , e ) 16
|
d
|
divide(multiply(16, 60), const_100)
|
multiply(n0,n1)|divide(#0,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 ?
|
"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 ) 11 , b ) 30 : 2 , c ) 28 , d ) 144 , e ) 88.2 %
|
e
|
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
|
sandy had $ 231 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ?
|
"let the money sandy took for shopping be x . 0.7 x = 231 x = 330 the answer is d ."
|
a ) 8 sec , b ) 380 , c ) $ 330 , d ) 465.60 $ , e ) 285
|
c
|
divide(231, divide(subtract(const_100, 30), const_100))
|
subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)|
|
gain
|
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 ?
|
"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 ) 3 , b ) 28 , c ) 4 , d ) 5.6 , e ) 501
|
c
|
divide(multiply(multiply(3, 8), 2), 12)
|
multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|
|
physics
|
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 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 ) 53.5 % , b ) 70 , c ) 720 , d ) 8 , e ) 12 hours
|
b
|
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
|
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 .
|
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 ) 6000 , b ) 50 years , c ) 6825 , d ) 0 , e ) 144 hours
|
a
|
multiply(multiply(6, const_10), const_100)
|
multiply(n2,const_10)|multiply(#0,const_100)
|
general
|
what is the number of integers from 1 to 100 ( inclusive ) that are divisible by neither 5 nor by 6 ?
|
"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 ) 58 , b ) $ 370,000 , c ) 320578 , d ) 991 , e ) 650
|
a
|
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
|
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 ?
|
looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c
|
a ) 25 % , b ) 75 kg , c ) 4 : 49 , d ) 15 , e ) 10.5
|
a
|
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
|
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 ?
|
"answer is c . x / x + 36 = 1 / 5 x = 9 16 - 9 = 7"
|
a ) 7 , b ) $ 44.33 , c ) 39 , d ) 16 , e ) 70 mph
|
a
|
multiply(divide(20, const_100), 20)
|
divide(n2,const_100)|multiply(n2,#0)|
|
gain
|
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 ?
|
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 ) 20 % , b ) 3 days , c ) 1 , d ) 60 , e ) 10.4
|
e
|
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 sum of the squares of three consecutive natural number is 2030 . what is the middle number ?
|
"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 ) 6400 , b ) 30 % , c ) 350 , d ) 26 , e ) s . 180
|
d
|
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
|
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 ?
|
"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 ) 7.5 sec , b ) 36 , c ) 120 , d ) $ 600 , e ) 9 / 49
|
a
|
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
|
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 ?
|
"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 ) 14 m , b ) 97 , c ) 4 : 49 , d ) 5 / 12 , e ) 5 %
|
e
|
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 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 ?
|
"age of the 15 th student = [ 15 * 15 - ( 14 * 5 + 16 * 9 ) ] = ( 225 - 214 ) = 11 years . answer : a"
|
a ) 6 , b ) 200 % , c ) 60 , d ) 11 years , e ) 982.14', '
|
d
|
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
|
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 ?
|
458 - 435 = 23 . answer is c .
|
a ) 40 , b ) 6 liters , c ) s 1770 , d ) 100 coins , e ) 23
|
e
|
subtract(458, 435)
|
subtract(n1,n0)
|
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 ?
|
"b 5 1 / 4 â € “ 1 / 20 = 1 / 5 = > 5"
|
a ) 28 / 55 , b ) 5.48 days , c ) - 14 , d ) 184 , e ) 5
|
e
|
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
|
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 ?
|
let x be the value of the item . 0.07 * ( x - 1000 ) = 109.90 x = 2570 the answer is d .
|
a ) 680 , b ) $ 2570 , c ) 600 , d ) 10 years , e ) 23 / 29
|
b
|
add(1000, divide(109.9, divide(7, const_100)))
|
divide(n0,const_100)|divide(n2,#0)|add(n1,#1)
|
general
|
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 ?
|
"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 ) 6.8 % , b ) 8 , c ) 2000 , d ) 210 , e ) 8.33 km
|
d
|
multiply(20, multiply(54, const_0_2778))
|
multiply(n2,const_0_2778)|multiply(n1,#0)|
|
physics
|
find the area , diameter = 11 m .
|
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 ) 22 , b ) 5625 , c ) 95.07 square meter', ' , d ) 30 , e ) 52
|
c
|
circle_area(divide(11, const_2))
|
divide(n0,const_2)|circle_area(#0)
|
physics
|
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 ?
|
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 ) 3 % , b ) 8 , c ) 1200 m ^ 2', ' , d ) rs . 240 , e ) 7
|
e
|
subtract(11, reminder(4, 5))
|
reminder(n3,n2)|subtract(n4,#0)
|
general
|
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 ?
|
"6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12400 = 3720 . answer : c"
|
a ) 3720 , b ) 11.5 , c ) 12 , d ) 1 , e ) $ 24.88
|
a
|
multiply(divide(6300, add(add(6300, 4200), 10500)), 12400)
|
add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|
|
gain
|
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 ?
|
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 ) 2.4 km , b ) 11.9 , c ) $ 306 , d ) 5 , e ) 10.8
|
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
|
if 40 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ?
|
"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 ) 1215 , b ) 8 hours , c ) 13 , d ) $ 37.8 , e ) 33.3 %
|
e
|
multiply(divide(subtract(40, 20), add(40, 20)), const_100)
|
add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)|
|
general
|
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 :
|
"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 ) rs . 740 , b ) 60 m , c ) 3 , d ) 18 , e ) 7.29
|
b
|
subtract(500, divide(multiply(subtract(500, 170), 4), 3))
|
subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)|
|
physics
|
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 ?
|
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 / 104 , b ) 2 , c ) 3 , d ) 1.5 days , e ) 1400
|
a
|
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
|
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 ?
|
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 ) 168 , c ) 64 , d ) 555681 , e ) 80
|
a
|
multiply(const_4, power(27, divide(const_1, const_3)))
|
divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)
|
geometry
|
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 ?
|
"total fish = x percentage of second catch = ( 2 / 40 ) * 100 = 5 % so , x * 5 % = 50 x = 1000 ans . a"
|
a ) 1300 , b ) 1000 , c ) 21 , d ) 2 , e ) 4 years
|
b
|
divide(40, divide(2, 40))
|
divide(n2,n1)|divide(n0,#0)|
|
gain
|
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 ?
|
"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 ) 6 days , b ) 1954404 , c ) 160 , d ) 64 , e ) 2
|
a
|
divide(multiply(6, 3), subtract(6, 3))
|
multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
|
physics
|
meena wrote all the numbers from 1 to 59,999 inclusive . how many digits did she write in total ?
|
"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 ) – 3 , b ) 9 , c ) 14 , d ) 6825 , e ) 288,889
|
e
|
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
|
if 3 < x < 6 < y < 9 , then what is the greatest possible positive integer difference of x and y ?
|
"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 ) $ 5 , b ) 21 , c ) 5 , d ) 252 , e ) 4
|
c
|
subtract(subtract(9, 3), const_1)
|
subtract(n2,n0)|subtract(#0,const_1)|
|
general
|
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 ?
|
"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 ) rs 66.66 , b ) 0 , c ) 15 , d ) 466 , e ) 1
|
d
|
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
|
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 : 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 ) 465.60 $ , b ) 73.43 , c ) 17 % , d ) 1 / 3 , e ) 24 days .
|
c
|
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
|
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
|
"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 ) 4 : 5 , b ) 16 , c ) 900 , d ) 3 , e ) 5
|
b
|
divide(multiply(20, const_4), add(const_4, const_1))
|
add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|
|
gain
|
convert the 19 / 36 m / s into kilometers per hour ?
|
"19 / 36 m / s = 19 / 36 * 18 / 5 = 19 / 10 = 1.9 kmph . answer : e"
|
a ) 870 , b ) 10 , c ) 45 , d ) 1.9 , e ) 9
|
d
|
multiply(const_3_6, divide(19, 36))
|
divide(n0,n1)|multiply(#0,const_3_6)|
|
physics
|
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 ?
|
"percentage error in calculated area = ( 5 + 5 + ( 5 ã — 5 ) / 100 ) % = 10.25 % answer : d"
|
a ) 567 , b ) 10.28 % , c ) 44 , d ) 10 kmph , e ) 4
|
b
|
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
|
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 ?
|
"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 ) 7.9 % , c ) 74 , d ) 0 , e ) 5
|
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
|
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 ?
|
"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 ) 77.77 meters , b ) 41 , c ) 10800 , d ) 6 , e ) 9
|
a
|
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
|
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 ?
|
# 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 ) 50 , b ) 1000 , c ) 50 % , d ) 7 / 17 , e ) 1200
|
d
|
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
|
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 ?
|
"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 ) 26 , c ) 4.0 % , d ) 55 , e ) 9600
|
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
|
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 ?
|
oa is definitely wrong . the answer should be e .
|
a ) 18 years , b ) 13 lb , c ) $ 1.75 , d ) rs . 1350 , e ) 1200
|
c
|
add(multiply(0.25, subtract(5, 1)), 0.25)
|
subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)|
|
general
|
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 % ?
|
"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 . 352 , b ) 1 / 15 , c ) 5 hours , d ) 85 , e ) 51
|
a
|
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
|
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 ?
|
"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 ) – 14 , b ) 600 , c ) 34 , d ) 4 , e ) $ 1200
|
e
|
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
|
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 ?
|
"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 ) 2000 , b ) 70 , c ) 32 % , d ) s . 180 , e ) 26
|
b
|
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
|
calculate the area of a triangle , if the sides of are 52 cm , 48 cm and 20 cm , what is its area ?
|
"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 ) 120 , c ) 7 , d ) − 60 , e ) 425
|
a
|
multiply(divide(48, const_2), 20)
|
divide(n1,const_2)|multiply(n2,#0)|
|
geometry
|
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 ?
|
"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 ) 226 , b ) 120 , c ) 20,160 , d ) 9.89 % , e ) $ 6000
|
c
|
multiply(multiply(2, 3), 2)
|
multiply(n3,n4)|multiply(n3,#0)|
|
general
|
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 ?
|
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 ) 4000 , b ) 75 , c ) 48 hours , d ) 7.36 kg , e ) 220010
|
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
|
how many positive integers less than 50 have a reminder 5 when divided by 7 ?
|
"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 ) rs . 98.56 , b ) ( 40 , c ) 64.3 , d ) 400,200 , e ) 7
|
e
|
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
|
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 : 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 ) rs 10123.20 , b ) 9990 , c ) 32 % , d ) $ 576 , e ) 1812.5
|
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 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 ?
|
"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 ) 10 sec , b ) 3 : 4 , c ) 64 , d ) 18 , e ) 14 %
|
b
|
divide(550, multiply(divide(33, 45), const_1000))
|
divide(n2,n3)|multiply(#0,const_1000)|divide(n0,#1)|
|
physics
|
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 ?
|
"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 ) 50', ' , b ) 12800 , c ) $ 4000 , d ) 642 , e ) 10
|
c
|
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
|
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 ?
|
"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 ) 16 , b ) 1 , c ) 25 , d ) 760 , e ) rs . 20
|
d
|
multiply(380, const_2)
|
multiply(n0,const_2)|
|
general
|
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 ?
|
"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 ) 215 , b ) 40 % , c ) 140 , d ) 24 % , e ) 2
|
b
|
multiply(divide(subtract(70, 42), 70), const_100)
|
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
|
gain
|
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 ?
|
"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 ) 240000 , b ) 20.5 , c ) 225 , d ) rs . 90 , e ) 25
|
e
|
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
|
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 ?
|
"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 ) 18 , b ) s . 180 , c ) 16 , d ) 480 , e ) none
|
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
|
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 ?
|
"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 ) 3 / 5 , b ) 15 sec , c ) 100 , d ) 3 / 10 , e ) 4608
|
a
|
divide(3, divide(75, const_10))
|
divide(n1,const_10)|divide(n0,#0)|
|
physics
|
andy solves problems 74 to 135 inclusive in a math exercise . how many problems does he solve ?
|
"135 - 74 + 1 = 62 ' b ' is the answer"
|
a ) 200 , b ) 1 / 27 , c ) 29 , d ) 62 , e ) 50 %
|
d
|
add(subtract(135, 74), const_1)
|
subtract(n1,n0)|add(#0,const_1)|
|
general
|
9.009 / 1.001
|
"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 ) 3 / 5 , b ) 7.05 , c ) 86 , d ) 9 , e ) 24
|
d
|
multiply(divide(9.009, 1.001), const_100)
|
divide(n0,n1)|multiply(#0,const_100)|
|
general
|
a number increased by 25 % gives 520 . the number is ?
|
"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 ) 4 , b ) 416 , c ) 12 , d ) 9 , e ) 991
|
b
|
divide(520, add(const_1, divide(25, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
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 ?
|
"x / 14 + x / 30 + x / 40 = 1 x = 7.7 days answer : b"
|
a ) 7.7 days , b ) 12 , c ) 0.225 , d ) 12 hours , e ) 57.5
|
a
|
add(subtract(subtract(14, subtract(40, 30)), 4), const_1)
|
subtract(n2,n1)|subtract(n0,#0)|subtract(#1,n3)|add(#2,const_1)|
|
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 ?
|
"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 ) 51 , b ) sqrt ( 2.88 / p , c ) 30 , d ) 35 , e ) 6.8 %
|
a
|
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
|
what is the sum of the multiples of 4 between 38 and 127 inclusive ?
|
"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 ) 95 , b ) 1534 , c ) 16 % , d ) 1804 , e ) 80 %
|
d
|
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
|
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 ?
|
"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 ) 14.5 , b ) 8 / 3 , c ) 6 , d ) 58.65 ft , e ) 12.5 m
|
a
|
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 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 ?
|
"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 ) 1.33 , b ) 12 , c ) rs . 5845 , d ) 11 sec , e ) four
|
d
|
divide(110, multiply(add(30, 6), const_0_2778))
|
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
|
physics
|
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 .
|
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 ) 49.9 kg , b ) 991 , c ) 762 , d ) 5.33 kmph , e ) 2
|
a
|
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
|
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 ?
|
"1 h - - - - - 5 ? - - - - - - 60 12 h rs = 20 + 21 = 41 t = 12 d = 41 * 12 = 492 answer : b"
|
a ) 492 , b ) 318 $ , c ) 9944 , d ) 6825 , e ) 16.5
|
a
|
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
|
how many positive integers less than 240 are multiple of 4 but not multiples of 6 ?
|
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 ) 192 , b ) 40 , c ) 8 , d ) 40 days , e ) 45
|
b
|
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
|
if w / x = 1 / 3 and w / y = 4 / 15 , then ( x + y ) / y =
|
"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 ) 1636 , b ) 5 , c ) 1800 , d ) 9 / 5 , e ) 35 %
|
d
|
add(divide(divide(4, 1), divide(15, 3)), const_1)
|
divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)|
|
general
|
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 ?
|
"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 ) $ 900 , b ) 6.8 % , c ) 29.8 % , d ) 14 : 00 , e ) 3 / 8
|
a
|
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
|
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 ?
|
"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 ) 46 , b ) 45 , c ) 6.6 kmph , d ) 1 / 15 , e ) 14
|
e
|
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 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 ?
|
"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 ) none of these , b ) 7.94 , c ) 1,120 , d ) 5 , e ) 16 hrs
|
a
|
subtract(265, multiply(divide(3, const_60), const_1000))
|
divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)|
|
physics
|
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 ?
|
6 flavours * 6 choices = 6 c 1 * 6 c 1 = 6 * 6 = 36 = d
|
a ) 15 , b ) 36 , c ) 69000 , d ) 50 , e ) 6 / 7
|
b
|
multiply(6, 6)
|
multiply(n0,n0)
|
general
|
the sum of three consecutive multiples of 3 is 117 . what is the largest number ?
|
"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 ) 6 , b ) 140 , c ) 26 , d ) 42 , e ) 3 .
|
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
|
in a zoo , the ratio of the number of cheetahs to the number 4 then what is the increase in the number of pandas ?
|
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 ) 30 , b ) 5 / 4 , c ) 12 , d ) 343 cc , e ) 16.36 %
|
c
|
subtract(multiply(4, 4), const_4)
|
multiply(n0,n0)|subtract(#0,const_4)
|
other
|
a train 250 m long running at 72 kmph crosses a platform in 50 sec . what is the length of the platform ?
|
"d = 72 * 5 / 18 = 50 = 1000 â € “ 250 = 750 m answer : d"
|
a ) 6 days , b ) 24 , c ) 750 m , d ) 10 % , e ) 12
|
c
|
subtract(multiply(50, multiply(72, const_0_2778)), 250)
|
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
|
physics
|
how many multiples of 3 are there between 5 and 59 , 5 and 59 inclusive ?
|
"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 ) 30 , 10 , b ) 18 , c ) 1540 , d ) 20 sec , e ) 2
|
b
|
add(divide(subtract(59, 5), 3), const_1)
|
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
|
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 .
|
"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 ) 3.5 , b ) 7 / 66 , c ) 12 / 7 , d ) 270 , e ) 450
|
c
|
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 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 ?
|
"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 ) 45 , b ) 1.33 , c ) 10.5 , d ) 7 , e ) 31
|
b
|
divide(divide(32.3, 3), multiply(4, 2))
|
divide(n3,n2)|multiply(n5,n6)|divide(#0,#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 ?
|
"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 ) 78 , b ) 200 , c ) 28 min , d ) 176 , e ) 1 : 1
|
a
|
divide(2496, multiply(power(const_2, const_4), const_2))
|
power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)|
|
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 ?
|
"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 . 352 , b ) 28 , c ) s 1770 , d ) 24 , e ) 9
|
c
|
multiply(4, divide(6490, add(add(4, 2), const_3)))
|
add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#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 ?
|
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 ) 140 , b ) 22 ½ days , c ) 625 , d ) 6 , e ) 220
|
c
|
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
|
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 ?
|
explanation : salary of the manager = ( 56 * 8800 - 55 * 8500 ) = 25300 answer : d
|
a ) 14 % , b ) 25300 , c ) 324 , d ) 9 , e ) 1
|
b
|
subtract(multiply(add(55, const_1), 8800), multiply(55, 8500))
|
add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)
|
general
|
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 ?
|
"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 ) 144 , b ) 6 , c ) 1,120 , d ) 1 / 27 , e ) 315
|
b
|
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
|
how many of the positive factors of 10 , 16 and how many common factors are there in numbers ?
|
"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 ) 19619 , b ) 1235 , c ) 3 , d ) 3.6 , e ) 1225
|
c
|
divide(16, 10)
|
divide(n1,n0)|
|
other
|
in a simultaneous throw of pair of dice . find the probability of getting the total more than 7
|
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 ) 3 : 2 , b ) 108 kmph , c ) 34 % , d ) 5 / 12 , e ) 18 feet
|
d
|
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
|
calculate the largest 6 digit number which is exactly divisible by 99 ?
|
"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 ) 999936 , b ) 170 m , c ) 1 / 3 , d ) 23 , e ) 145
|
a
|
multiply(add(const_100, const_2), 99)
|
add(const_100,const_2)|multiply(n1,#0)|
|
general
|
a green grocer received a boxful of tomatoes and on opening the box found that several had gone bad . he then counted them up so that he could make a formal complaint and found that 68 were mouldy , which was 16 per cent of the total contents of the box . how many tomatoes were in the box ?
|
b 425 ( 68 ã · 16 ) ã — 100
|
a ) 425 , b ) 94 kmph , c ) 33.33 % , d ) 30 , e ) 720
|
a
|
subtract(multiply(multiply(68, const_4), const_2), const_100)
|
multiply(n0,const_4)|multiply(#0,const_2)|subtract(#1,const_100)
|
general
|
the water level in a reservoir has been dropping at the rate of 12 inches per day . exactly 5 days ago , the water level was at w inches . what will be the water level exactly 4 days from now if the rate at which the level is dropping remains the same ?
|
drop = 12 inches / day 5 days ago = w , means now it ' s equal w - 60 and in 4 days = w - 60 - 48 = w - 108 answer a
|
a ) 23 % . , b ) 1453 , c ) w − 108 , d ) 20 , e ) 1000
|
c
|
multiply(12, divide(4, 5))
|
divide(n2,n1)|multiply(n0,#0)
|
gain
|
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