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a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 11 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ?
"ans : 385 7 c 1 * 11 c 2 answer e )"
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 385
e
multiply(multiply(11, 3), 7)
multiply(n3,n5)|multiply(n1,#0)|
other
in the year 1990 there are 5000 men 3000 women 2000 boys . in 1994 men are increased by 20 % women are increased by ratio of boys and women ?
total population in year 1990 = 5000 + 3000 + 2000 = 10000 total population in year 1994 = ( ( 5000 * 120 / 100 ) + ( 3000 + ( 3000 * 2 / 3 ) ) + ( 2000 ) ) = 13000 answer : c
a ) 11000 , b ) 12000 , c ) 13000 , d ) 14000 , e ) 15000
c
add(add(multiply(5000, add(const_1, divide(20, const_100))), multiply(3000, add(const_1, divide(2000, 3000)))), 2000)
divide(n5,const_100)|divide(n3,n2)|add(#0,const_1)|add(#1,const_1)|multiply(n1,#2)|multiply(n2,#3)|add(#4,#5)|add(n3,#6)
other
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , the how old is b
"let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 x = 5 . hence , b ' s age = 2 x = 10 years answer : a"
a ) 10 , b ) 8 , c ) 6 , d ) 5 , e ) 7
a
divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1))
add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)|
general
what is the sum of the multiples of 4 between 13 and 125 inclusive ?
"the first multiple of 12 between 13 and 125 is 16 while the last is 124 total number of terms is given by 124 = 16 + ( n - 1 ) 4 4 n - 4 = 108 n = 112 / 4 = 28 or , ( 124 - 16 ) / 4 + 1 = 28 so , we have 28 multiples of 12 between 13 and 125 sum = 14 ( 2 * 16 + ( 27 ) 4 ) = 14 ( 140 ) = 1960 answer : b"
a ) 1,890 , b ) 1,960 , c ) 2,200 , d ) 3,780 , e ) 4,400
b
multiply(divide(add(subtract(125, const_3), add(13, const_2)), const_2), add(divide(subtract(subtract(125, const_3), add(13, 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
the volume of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate s , in cubic meters per minutes , at which the water is flowing into the swimming pool ?
"the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e"
a ) 0.125 , b ) 0.25 , c ) 0.5 , d ) 0.75 , e ) not enough information to determine the rate
e
divide(840, 0.5)
divide(n0,n1)|
geometry
four horses are tethered at 4 corners of a square field of side 70 metres so that they just can not reach one another . the area left ungrazed by the horses is :
area ungrazed is given by total area - 4 * area grazed by each horse = 70 * 70 - 4 * ( 90 / 360 ) * pi * ( 70 / 2 ) ^ 2 as the angle made by the horse is 90 degree , so applying the area of the sector , = theta / 360 * pi * radius ^ 2 above = 70 * 70 - pi * ( 70 / 2 ) * ( 70 / 2 ) = 70 * 70 { 1 - pi / 4 } = 70 * 70 { 6 / ( 7 * 4 ) } , expanding pi = 22 / 7 = ( 70 * 70 * 6 ) / ( 7 * 4 ) = 1050 sq m answer : a
['a ) 1050 sq . m', 'b ) 3850 sq . m', 'c ) 950 sq . m', 'd ) 1075 sq . m', 'e ) 1065 sq . m']
a
subtract(square_area(70), circle_area(divide(70, const_2)))
divide(n1,const_2)|square_area(n1)|circle_area(#0)|subtract(#1,#2)
geometry
a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 38 . find the profit percentage ?
l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 190 ( 30 * 5 ) profit percentage = ( 190 - 100 ) / 100 * 100 = 90 % answer : b
a ) 80 % , b ) 90 % , c ) 59 % , d ) 40 % , e ) 53 %
b
subtract(multiply(38, add(const_4, const_1)), multiply(25, const_4))
add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)
gain
redo ’ s manufacturing costs for sets of horseshoes include a $ 10000 initial outlay , and $ 20 per set . they can sell the sets $ 50 . if profit is revenue from sales minus manufacturing costs , and the company producessells 500 sets of horseshoes , what was their profit ?
total manufacturing cost = 10000 + 500 * 20 = 20000 total selling cost = 500 * 50 = 25000 profit = 25000 - 20000 = 5000 answer : d
a ) $ 4500 , b ) $ 3500 , c ) $ 5500 , d ) $ 5000 , e ) $ 6300
d
subtract(multiply(subtract(50, 20), 500), 10000)
subtract(n2,n1)|multiply(n3,#0)|subtract(#1,n0)
gain
from below option 50 is divisible by which one ?
"50 / 2 = 25 b"
a ) 3 , b ) 2 , c ) 7 , d ) 8 , e ) 9
b
sqrt(50)
sqrt(n0)|
general
in n is a positive integer less than 200 , and 24 n / 60 is an integer , then n has how many different positive prime factors ?
"( a ) . 24 n / 60 must be an integer . = > 2 n / 5 must be an integer . hence n must be a multiple of 5 . = > n has 1 different prime integers ."
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 8
a
add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2)
add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)|
general
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 4 cups of tea . last week mo drank a total of 42 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how many rainy days were there last week ?
"t = the number of cups of tea c = the number of cups of hot chocolate t + c = 42 t - c = 14 - > t = 28 . c = 14 . mo drinks 4 cups of tea a day then number of days that are not rainy = 28 / 4 = 7 so number of rainy days = 7 - 7 = 0 a is the answer ."
a ) 0 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
subtract(add(const_4, 4), divide(divide(add(42, 14), const_2), 4))
add(n0,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)|
general
469157 * 9999
"explanation : 469157 * ( 10000 - 1 ) = 4691570000 - 469157 = 4691100843 option c"
a ) 1691100843 , b ) 4591100843 , c ) 4691100843 , d ) 3691100843 , e ) none of these
c
multiply(divide(469157, 9999), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
what is the largest 4 digit number exactly divisible by 72 ?
"largest 4 digit number = 9999 9999 ÷ 72 = 138 , remainder = 63 hence largest 4 digit number exactly divisible by 88 = 9999 - 63 = 9936 answer : e"
a ) 4676 , b ) 4678 , c ) 8888 , d ) 9504 , e ) 9936
e
multiply(floor(divide(power(const_10, 4), 72)), 72)
power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)|
general
apple costs l rupees per kilogram for first 30 kgs and q rupees per kilogram for each additional kilogram . if the price of 33 kilograms is 360 and for 36 kgs of apples is 420 then the cost of first 25 kgs of apples is
"ans : by framing equations we get 30 l + 3 q = 360 30 l + 6 q = 420 eliminate q by multiplying the first equation by 2 and subtracting second equation from the first then we get l = 10 cost of 10 kgs of apples = 25 x 10 = 250 answer : b"
a ) 100 , b ) 250 , c ) 125 , d ) 110 , e ) 115
b
multiply(divide(subtract(360, multiply(subtract(33, 30), divide(subtract(420, 360), subtract(36, 33)))), 30), 25)
subtract(n4,n2)|subtract(n3,n1)|subtract(n1,n0)|divide(#0,#1)|multiply(#3,#2)|subtract(n2,#4)|divide(#5,n0)|multiply(n5,#6)|
other
if x = 3 ^ 27 and x ^ x = 3 ^ k , what is k ?
"solution : we know that x = 3 ^ 27 which implies x ^ x = ( 3 ^ 27 ) ^ ( 3 ^ 27 ) = 3 ^ ( 27 * 3 ^ 27 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 3 ^ ( 3 ^ 3 * 3 ^ 27 ) = 3 ^ ( 3 ^ ( 3 + 27 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 3 ^ ( 3 ^ 30 ) given that x ^ x = 3 ^ k so 3 ^ ( 3 ^ 30 ) = 3 ^ k since the base is same the exponent will also be same therefore k = 3 ^ 30 answer : d"
a ) 3 ^ 9 , b ) 3 ^ 12 , c ) 3 ^ 27 , d ) 3 ^ 30 , e ) 3 ^ 33
d
multiply(power(3, 27), 27)
power(n0,n1)|multiply(n1,#0)|
general
find the least number which when divided by 5,6 , 7,8 leaves a remainder 3 but when divided by 9 leaves no remainder
"l . c . m of 5,6 , 7,8 = 840 required number is of the form of 840 k + 3 least value of k for which ( 840 k + 3 ) is divided by 9 is k = 2 required number = ( 840 * 2 + 3 ) = 1683 answer ( d )"
a ) 9632 , b ) 7896 , c ) 8741 , d ) 1683 , e ) 8523
d
add(3, lcm(5,6, 7,8))
lcm(n0,n1)|add(n2,#0)|
general
the ratio of the radius of two circles is 3 : 4 , and then the ratio of their areas is ?
"r 1 : r 2 = 3 : 4 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 9 : 16 answer : b"
a ) 1 : 7 , b ) 9 : 16 , c ) 1 : 9 , d ) 3 : 7 , e ) 3 : 4
b
divide(circle_area(3), circle_area(4))
circle_area(n0)|circle_area(n1)|divide(#0,#1)|
geometry
evaluate : 30 - | - x + 6 | for x = 10
substitute x by 10 in the given expression and evaluate 30 - | - ( 10 ) + 6 | = 30 - | - 10 + 6 | = 30 - | - 4 | = 30 - 4 = 26 correct answer b ) 26
a ) 16 , b ) 26 , c ) 36 , d ) 46 , e ) 56
b
subtract(30, negate(add(negate(10), 6)))
negate(n2)|add(n1,#0)|negate(#1)|subtract(n0,#2)
general
in a partnership , a invests 1 ⁄ 6 of the capital for 1 ⁄ 6 of the time , b invests 1 ⁄ 3 of the capital for 1 ⁄ 3 of the time and c , the rest of the capital for whole time . find a ’ s share of the total profit of 2300
remaining capital = 1 - ( 1 ⁄ 6 + 1 ⁄ 3 ) = 1 ⁄ 2 ratio of their profit = 1 ⁄ 6 × [ 1 ⁄ 6 × 12 ] : 1 ⁄ 3 × [ 1 ⁄ 3 × 12 ] : 1 ⁄ 2 × 12 = 1 ⁄ 3 : 4 ⁄ 3 : 6 = 1 : 4 : 18 ∴ a ' s share = 1 / 1 + 4 + 18 × 2300 = 100 answer a
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) none of these
a
multiply(inverse(add(multiply(6, 3), add(const_1, const_4))), 2300)
add(const_1,const_4)|multiply(n1,n5)|add(#0,#1)|inverse(#2)|multiply(n8,#3)
gain
a shopkeeper has 280 kg of apples . he sells 40 % of these at 20 % profit and remaining 50 % at 30 % profit . find his % profit on total .
"if the total quantity was 100 then 40 x 20 % + 50 x 30 % = 23 this profit will remain same for any total quantity unless the % of products remains the same . hence ' a ' is the answer"
a ) 23 % , b ) 25 % , c ) 26 % , d ) 28 % , e ) 35 %
a
divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 50), const_100), divide(add(const_100, 30), const_100))), 280), const_100), 280)
add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)|
gain
car x began traveling at an average speed of 35 miles per hour . after 48 minutes , car y began traveling at an average speed of 39 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
"in 48 minutes , car x travels 28 miles . car y gains 4 miles each hour , so it takes 7 hours to catch car x . in 7 hours , car x travels 245 miles . the answer is e ."
a ) 105 , b ) 140 , c ) 175 , d ) 210 , e ) 245
e
multiply(35, divide(multiply(divide(48, const_60), 35), subtract(39, 35)))
divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)|
physics
how many 4 - digit positive integers are there in which all 4 digits are even ?
"positive integers - 2 , 4,6 , 8,0 let the integers of a four digit positive number be abcd a can take four values ( 2,4 , 6,8 ) b can take five values ( 0 , 2,4 , 6,8 ) c can take five values ( 0 , 2,4 , 6,8 ) d can take five values ( 0 , 2,4 , 6,8 ) the total is 5 * 5 * 5 * 4 the answer according to me is 500 answer : c"
a ) 625 , b ) 600 , c ) 500 , d ) 400 , e ) 256
c
multiply(multiply(add(4, 4), add(4, 4)), multiply(add(4, 4), multiply(4, 4)))
add(n1,n0)|add(n0,n0)|multiply(n1,n1)|multiply(#0,#0)|multiply(#1,#2)|multiply(#3,#4)|
general
benny goes to the market for buying some apples to be distributed between her 9 kids equally . she takes 360 dollars with her . the cost of each apple is 4 dollars . how many apples does she buy to share them equally between her eighteen kids ?
cost of each apple = 4 dollars apples that benny can buy with the amount she has = 360 / 4 = 90 . apples that each kid gets evenly = 90 / 18 = 5 apples . so the answer is c = 5
a ) 8 , b ) 9 , c ) 5 , d ) 7 , e ) 10
c
divide(divide(360, 4), multiply(9, const_2))
divide(n1,n2)|multiply(n0,const_2)|divide(#0,#1)
general
it takes nine minutes to load a certain video on a cellphone , and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates , how many seconds would it take them to load the video , rounded to the nearest hundredth ?
the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 9 ) = 1 / 540 of the video per second . the combined rate is 1 / 15 + 1 / 540 = 37 / 540 of the video per second . the time required to load the video is 540 / 37 = 14.59 seconds . the answer is d .
a ) 13.58 , b ) 13.87 , c ) 14.24 , d ) 14.59 , e ) 14.85
d
subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))
add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)
physics
a sum of money invested at compound interest to rs . 800 in 3 years and to rs 820 in 4 years . the rate on interest per annum is .
explanation : s . i . on rs 800 for 1 year = 20 rate = ( 100 * 20 ) / ( 800 * 1 ) = 2.5 % answer : c
a ) 0.4 % , b ) 5.0 % , c ) 2.5 % , d ) 7 % , e ) 8 %
c
divide(multiply(const_100, subtract(820, 800)), 800)
subtract(n2,n0)|multiply(#0,const_100)|divide(#1,n0)
gain
if two positive numbers are in the ratio 1 / 9 : 1 / 7 , then by what percent is the second number more than the first ?
"given ratio = 1 / 9 : 1 / 7 = 7 : 9 let first number be 7 x and the second number be 9 x . the second number is more than first number by 2 x . required percentage = 2 x / 7 x * 100 = 28.6 % . answer : d"
a ) 67 % . , b ) 70 % . , c ) 60 % . , d ) 28.6 % . , e ) 80 % .
d
multiply(divide(1, 7), const_100)
divide(n0,n3)|multiply(#0,const_100)|
general
find compound interest on rs . 7500 at 4 % per annum for 2 years , compounded annually
explanation : please apply the formula amount = p ( 1 + r 100 ) nc . i . = amount - p answer : d
a ) rs 312 , b ) rs 412 , c ) rs 512 , d ) rs 612 , e ) none of these
d
subtract(add(add(7500, divide(multiply(7500, const_4), const_100)), divide(multiply(add(7500, divide(multiply(7500, const_4), const_100)), 4), const_100)), 7500)
multiply(n0,const_4)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)
gain
the front wheels of a wagon are 2 π feet in circumference and the rear wheels are 3 π feet in circumference . when the front wheels have made 10 more revolutions than the rear wheels , how many feet has the wagon travelled ?
solution let the rear wheel make x revolutions . then , the front wheel makes ( x + 10 ) revolutions . ( x + 10 ) x 3 π = x × 2 π ‹ = › 3 x + 30 = 2 x ‹ = › x = 30 . distance travelled by the wagon = ( 2 π x 30 ) ft ‹ = › ( 60 π ) ft . answer c
a ) 30 π , b ) 45 π , c ) 60 π , d ) 90 π , e ) none
c
divide(multiply(multiply(multiply(2, 10), 3), multiply(subtract(const_12, const_1), 2)), add(const_3, const_4))
add(const_3,const_4)|multiply(n0,n2)|subtract(const_12,const_1)|multiply(n1,#1)|multiply(n0,#2)|multiply(#3,#4)|divide(#5,#0)
geometry
lcm of two numbers is 7700 and hcf is 11 . if one number is 308 then other number is
option d
a ) 269 , b ) 285 , c ) 300 , d ) 275 , e ) none of these
d
divide(multiply(7700, 11), 308)
multiply(n0,n1)|divide(#0,n2)
physics
a car is running at a speed of 80 kmph . what distance will it cover in 20 sec ?
speed = 80 kmph = 80 * 5 / 18 = 22 m / s distance covered in 20 sec = 22 * 10 = 440 m answer is b
a ) 100 m , b ) 440 m , c ) 180 m , d ) 200 m , e ) 250 m
b
multiply(divide(80, const_3_6), 20)
divide(n0,const_3_6)|multiply(n1,#0)
physics
a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , a receives
"solution : let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 . = > 3 x = 36000 . = > x = 12000 . a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . so a ' s share = rs . ( 35000 x 21 / 50 ) = rs . 14,700 . answer : option d"
a ) rs . 8400 , b ) rs . 11,900 , c ) rs . 13,600 , d ) rs . 14,700 , e ) none
d
subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1)
add(n1,n2)|add(const_3,const_4)|multiply(n2,const_10)|multiply(n2,#1)|subtract(#2,n2)|subtract(#4,#0)|divide(#5,const_3)|add(#0,#6)|divide(#7,#2)|multiply(#8,#3)|divide(#9,const_1000)|floor(#10)|subtract(#11,const_1)|
general
the radius of a cylindrical vessel is 7 cm and height is 2 cm . find the whole surface of the cylinder ?
"r = 7 h = 2 2 π r ( h + r ) = 2 * 22 / 7 * 7 ( 9 ) = 396 answer : b"
a ) 281 , b ) 396 , c ) 440 , d ) 767 , e ) 1981
b
surface_cylinder(7, 2)
surface_cylinder(n0,n1)|
geometry
a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number r of tickets sold by the male members of the association
"given that , total average r sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ."
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66
c
subtract(multiply(66, add(1, 2)), multiply(70, 2))
add(n2,n3)|multiply(n1,n3)|multiply(n0,#0)|subtract(#2,#1)|
general
a train 880 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ?
"speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 880 + x ) / 60 = 65 / 3 x = 420 m . answer : option d"
a ) 510 , b ) 540 , c ) 500 , d ) 420 , e ) 589
d
divide(880, multiply(subtract(78, 1), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
if x is less than y by 50 % then y exceed x by :
"using formula ( x / ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 25 % ) = > 50 ( 100 - 50 ) * 100 = 100 % answer : b"
a ) 33.33 % , b ) 100 % , c ) 75 % , d ) 66.66 % , e ) none of these
b
multiply(subtract(divide(const_100, subtract(const_100, 50)), const_1), const_100)
subtract(const_100,n0)|divide(const_100,#0)|subtract(#1,const_1)|multiply(#2,const_100)|
general
arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 30 % profit ?
"explanation : c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 130 % of rs . 630 = 130 / 100 * 630 = rs . 819 . s . p . per kg = 819 / 50 = rs . 16.38 = 16.30 . answer : d"
a ) 66.3 , b ) 76.3 , c ) 86.3 , d ) 16.3 , e ) 36.3
d
divide(add(add(multiply(30, 11.50), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.50), multiply(20, 14.25)), const_100), 30)), add(30, 20))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,const_100)|multiply(n4,#4)|add(#3,#5)|divide(#6,#0)|
gain
what was the principal , if at 5 % per annum the interest after 10 years amounted to rs . 3100 less than the sum lent ?
"p - 2336 = ( p * 10 * 5 ) / 100 p = 6200 answer : a"
a ) 6200 , b ) 3000 , c ) 3000 , d ) 2000 , e ) 1000
a
divide(3100, subtract(const_1, divide(multiply(5, 10), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
gain
the list price of an article is rs . 68 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?
"68 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 8.23 % answer : e"
a ) 3.23 % , b ) 4.23 % , c ) 5.23 % , d ) 7.23 % , e ) 8.23 %
e
multiply(divide(subtract(subtract(68, multiply(68, divide(10, const_100))), 56.16), subtract(68, multiply(68, divide(10, const_100)))), const_100)
divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)|
gain
the cost of the paint is rs . 36.50 per kg . if 1 kg of paint covers 16 square feet , how much will it cost to paint outside of a cube having 8 feet each side .
"explanation : we will first calculate the surface area of cube , then we will calculate the quantity of paint required to get answer . here we go , surface area = 6 a 2 = 6 ∗ 8 ( 2 ) = 384 sq feet quantity required = 38416 = 24 kg cost of painting = 36.50 ∗ 24 = rs . 876 option c"
a ) rs . 850 , b ) rs . 860 , c ) rs . 876 , d ) rs . 886 , e ) none of these
c
multiply(divide(surface_cube(8), 16), 36.50)
surface_cube(n3)|divide(#0,n2)|multiply(n0,#1)|
geometry
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many w hours will it take if jim types the report and mary edits it immediately after he is done ?
"break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many w hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a ."
a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8
a
add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12))))
divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|
physics
a scuba diver descends at a rate of 35 feet per minute . a diver dive from a ship to search for a lost ship at the depth of 3500 feet below the sea level . . how long will he take to reach the ship ?
"time taken to reach = 3500 / 35 = 100 minutes answer : c"
a ) 160 minutes , b ) 120 minutes , c ) 100 minutes , d ) 76 minutes , e ) 77 minutes
c
divide(3500, 35)
divide(n1,n0)|
gain
a snail , climbing a 42 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?
"total transaction in two days = 4 - 2 = 2 feet in 38 days it will climb 38 feet on the 39 th day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 39 d"
a ) 20 , b ) 26 , c ) 32 , d ) 39 , e ) 51
d
subtract(42, 4)
subtract(n0,n1)|
physics
at what rate percent on simple interest will rs . 900 amount to rs . 950 in 5 years ?
"50 = ( 900 * 5 * r ) / 100 r = 1.11 % answer : a"
a ) 1.11 % , b ) 5.93 % , c ) 4.33 % , d ) 5.33 % , e ) 6.33 %
a
multiply(divide(divide(subtract(950, 900), 900), 5), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
if the price of sugar rises from rs . 10 per kg to rs . 12 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by
"sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x × 12 = x = 83 kg . ∴ reduction in consumption = 17 % . answer a"
a ) 17 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) none
a
multiply(subtract(const_1, divide(multiply(const_1, 10), 12)), const_100)
multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)|
general
initially , the men and women in a room were in the ratio of 4 : 5 . then , 2 men entered the room and 3 women left the room . then , the number of women doubled . now there are 14 men in the room . how many e women are currently in the room ?
"the number of women doubled means that they have become 24 from 12 . . and we have to tell the current strength so 24 is the answer . . let the number be 4 x and 5 x . . given 4 x + 2 = 14 . . so x = 3 . . women number = 5 * 3 - 3 = 12 , then doubled = 24 . . ans d"
a ) 12 , b ) 14 , c ) 15 , d ) 24 , e ) 36
d
multiply(2, subtract(divide(multiply(5, subtract(14, 2)), 4), 3))
subtract(n4,n2)|multiply(n1,#0)|divide(#1,n0)|subtract(#2,n3)|multiply(n2,#3)|
other
find the compound ratio of ( 2 : 3 ) , ( 3 : 4 ) and ( 1 : 4 ) is
"required ratio = 2 / 3 * 3 / 4 * 1 / 4 = 1 / 8 = 1 : 8 answer is c"
a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 8 , d ) 4 : 5 , e ) 3 : 2
c
multiply(divide(2, const_3.0), multiply(divide(2, 3), divide(3, 3)))
divide(n0,n3)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)|
other
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 16 households had a both a car and a bike and 44 had a car , how many had bike only ?
"{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 16 + 11 - - > { bike } = 51 - - > # those who have bike only is { bike } - { both } = 51 - 16 = 35 . answer : b ."
a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10
b
subtract(subtract(add(subtract(90, 11), 16), 44), 16)
subtract(n0,n1)|add(n2,#0)|subtract(#1,n3)|subtract(#2,n2)|
other
a bag marked at $ 150 is sold for $ 120 . the rate of discount is ?
"rate of discount = 30 / 150 * 100 = 20 % answer is c"
a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 45 %
c
multiply(divide(subtract(150, 120), 150), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
a man swims downstream 70 km and upstream 30 km taking 10 hours each time ; what is the speed of the current ?
"70 - - - 10 ds = 7 ? - - - - 1 30 - - - - 10 us = 3 ? - - - - 1 s = ? s = ( 7 - 3 ) / 2 = 2 answer : c"
a ) 1 , b ) 4 , c ) 2 , d ) 6 , e ) 8
c
divide(add(divide(30, 10), divide(70, 10)), const_2)
divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|
physics
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 8 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ?
"ans : 196 7 c 1 * 8 c 2 answer d )"
a ) 42 , b ) 70 , c ) 140 , d ) 196 , e ) 315
d
multiply(multiply(8, 3), 7)
multiply(n3,n5)|multiply(n1,#0)|
other
a certain store sells all maps at one price and all books at another price . on monday the store sold 12 maps and 10 books for a total of $ 38.00 , and on tuesday the store sold 12 maps and 9 books for a total of $ 30.00 . at this store , how much less does a map sell for than a book ?
"12 x + 10 y = 38 12 x + 9 y = 30 subtracting 1 from 2 y = 8 therefore x = 3.5 difference in price = 4.5 e"
a ) $ 0.25 , b ) $ 0.50 , c ) $ 0.75 , d ) $ 1.00 , e ) $ 4.5
e
divide(subtract(30.00, negate(multiply(9, divide(subtract(multiply(38.00, const_4), multiply(30.00, const_2)), subtract(multiply(10, const_4), multiply(9, const_2)))))), 12)
multiply(n2,const_4)|multiply(n5,const_2)|multiply(n1,const_4)|multiply(n4,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n4,#6)|negate(#7)|subtract(n5,#8)|divide(#9,n3)|
general
if an object travels at three feet per second , how many feet does it travel in one hour ?
explanation : if an object travels at 2 feet per second it covers 3 x 60 feet in one minute , and 3 x 60 x 60 feet in one hour . answer = 10800 answer : b ) 10800
a ) 3488 , b ) 10800 , c ) 12788 , d ) 1800 , e ) 2881
b
multiply(multiply(const_3, const_60), const_60)
multiply(const_3,const_60)|multiply(#0,const_60)
physics
the length of a rectangle is twice its breadth . if its length is decreased by 5 cm and breadth is increased by 5 cm , the area of the rectangle is increased by 55 sq . cm . find the length of the rectangle .
"explanation : let breadth = x . then , length = 2 x . then , ( 2 x - 5 ) ( x + 5 ) - 2 x * x = 55 = > 5 x - 25 = 55 = > x = 16 . length of the rectangle = 16 cm . answer : option a"
a ) 16 cm , b ) 18 cm , c ) 15 cm , d ) 20 cm , e ) 23 cm
a
divide(const_100, const_3)
divide(const_100,const_3)|
geometry
a cylindrical container with 6 meters diameter and a height of 8 meters is filled to capacity with water . if the water is then used to fill several smaller cylinders ( 2 meters radius and 5 meters height ) , how many smaller cylinders can be filled to capacity before the larger cylinder becomes empty ?
calculate the volume of the larger cylinder and divide it by the volume of the smaller cylinder . volume of cylinder = π r 2 h larger cylinder volume = 226.19 smaller cylinder volume = 62.83 therefore the number of smaller cylinders can be filled to capacity = 226.19 / 62.83 = 3.6 answer is c only 3 smaller cylinders can be filled to capacity .
['a ) 5', 'b ) 4', 'c ) 3', 'd ) 2', 'e ) 1']
c
subtract(divide(volume_cylinder(divide(6, const_2), 8), volume_cylinder(2, 5)), divide(add(const_4, const_2), const_10))
add(const_2,const_4)|divide(n0,const_2)|volume_cylinder(n2,n3)|divide(#0,const_10)|volume_cylinder(#1,n1)|divide(#4,#2)|subtract(#5,#3)
geometry
four circular cardboard pieces , each of radius 7 cm are placed in such a way that each piece touches two other pieces . the area of the space encosed by the four pieces is. four circular cardboard pieces , each of radius 7 cm are placed in such a way that each piece touches two other pieces . the area of the space encosed by the four pieces is
required area = 14 * 14 - ( 4 * 1 / 4 * 22 / 7 * 7 * 7 ) sq cm = 196 - 154 = 42 sq cm . answer : c
['a ) 12', 'b ) 32', 'c ) 42', 'd ) 52', 'e ) 58']
c
subtract(power(multiply(7, const_2), const_2), multiply(power(7, const_2), const_pi))
multiply(n0,const_2)|power(n0,const_2)|multiply(#1,const_pi)|power(#0,const_2)|subtract(#3,#2)
geometry
the roof of an apartment building is rectangular and its length is 5 times longer than its width . if the area of the roof is 720 feet squared , what is the difference between the length and the width of the roof ?
"answer is e : 48 let w be the width , so length is 5 w . therefore : w * 5 w = 720 , solving for , w = 12 , so 5 w - w = 4 w = 4 * 12 = 48"
a ) 38 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 48 .
e
subtract(multiply(sqrt(divide(720, 5)), 5), sqrt(divide(720, 5)))
divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)|
geometry
jaclyn buys $ 10 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter
explanation : i = ( p x r x t ) / 100 = 10000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 237.5 answer : a
a ) 237.5 , b ) 234 , c ) 289.5 , d ) 345 , e ) none of these
a
divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4)
multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_4)
gain
two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 40 kmph respectively . in what time will they cross each other completely ?
"explanation : d = 250 m + 250 m = 500 m rs = 80 + 40 = 120 * 5 / 18 = 100 / 3 t = 500 * 3 / 100 = 15 sec answer : option a"
a ) 15 sec , b ) 19 sec , c ) 12 sec , d ) 10 sec , e ) 11 sec
a
divide(250, multiply(80, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
at garage sale , all of the prices of the items sold were different . if the price of a radio sold at the garage sale was both the 15 th highest price and the 25 th lowest price among the prices of the items sold , how many items were sold at the garage sale ?
"14 + 24 + 1 = 39 answer : d"
a ) 33 , b ) 34 , c ) 35 , d ) 39 , e ) 37
d
subtract(add(15, 25), const_1)
add(n0,n1)|subtract(#0,const_1)|
other
in a division sum , the quotient is 20 , the divisor 66 and the remainder 55 , find the dividend ?
"explanation : 20 * 66 + 55 = 1375 answer : a"
a ) 1375 , b ) 1376 , c ) 1875 , d ) 1365 , e ) 1345
a
add(multiply(20, 66), 55)
multiply(n0,n1)|add(n2,#0)|
general
a and b are two partially filled buckets of water . if 6 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 6 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ?
let bucket a be a and bucket b be b scenario 1 a - 6 = 1 / 3 ( b + 6 ) - - - - > 3 a - 18 = b + 6 scenario 2 b - 6 = 1 / 2 ( a + 6 ) - - - - - > 2 b - 12 = a + 6 from scenario 1 , b = 3 a - 24 substitute b with this information in stmt 2 2 ( 3 a - 24 ) - 12 = a + 6 - - - - - - > 6 a - 48 - 12 = a + 6 - - - - - - > 6 a - a = 60 + 6 - - - > 5 a = 66 a = 66 / 5 , answer choice a
a ) 66 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23
a
divide(multiply(add(add(6, const_3), const_2), divide(6, const_2)), add(const_2, divide(const_1, const_2)))
add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)
general
what is the total cost of 2 sandwiches at $ 2.49 each and 4 sodas at $ 1.87 each ?
"answer = e 2 * 2.49 + 4 * 1.87 = 2 ( 2.50 - 0.01 ) + 4 ( 2.00 - 0.13 ) = 5 + 8 - 0.02 - 0.52 = 13 - 0.54 = 12.46"
a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.46 , d ) $ 10.08 , e ) $ 12.46
e
add(multiply(2, 2.49), multiply(4, 1.87))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
general
the sum of ages of 4 children born 1 years different each is 38 years . what is the age of the elder child ?
"let the ages of children be x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) years . then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) = 38 4 x = 32 x = 8 . x + 3 = 8 + 3 = 11 answer : b"
a ) 8 , b ) 11 , c ) 10 , d ) 16 , e ) 17
b
divide(add(add(add(add(const_2.0, const_4), add(1, const_4)), add(const_4, const_4)), 38), 4)
add(const_2.0,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)|
general
what is the positive integer that can be added by 595 to make it a perfect square ?
596 is added to a number that gives a perfect square nearest perfect square is 625 . so , 625 - 596 = 30 30 = 2 * 3 * 5 option b is answer
['a ) 2 * 3 * 6', 'b ) 3 * 2 * 5', 'c ) 5 * 3 * 3', 'd ) 2 * 3', 'e ) 3 * 5 * 4']
b
subtract(power(add(floor(sqrt(595)), const_1), const_2), 595)
sqrt(n0)|floor(#0)|add(#1,const_1)|power(#2,const_2)|subtract(#3,n0)
geometry
after a storm deposits 110 billion gallons of water into the city reservoir , the reservoir is 60 % full . if the original contents of the reservoir totaled 220 billion gallons , the reservoir was approximately what percentage full before the storm ?
when the storm deposited 110 billion gallons , volume of water in the reservoir = 220 + 110 = 360 billion gallons if this is only 60 % of the capacity of the reservoir , the total capacity of the reservoir = 330 / 0.6 = 550 billion gallons therefore percentage of reservoir that was full before the storm = ( 220 / 550 ) * 100 = 40 % option a
a ) 40 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 %
a
multiply(divide(220, divide(add(110, 220), divide(60, const_100))), const_100)
add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|divide(n2,#2)|multiply(#3,const_100)
general
( 6 ) 6.5 × ( 36 ) 4.5 ÷ ( 216 ) 4.5 = ( 6 ) ?
"explanation : ( 6 ) 6.5 × ( 36 ) 4.5 ÷ ( 216 ) 4.5 = ( 6 ) 6.5 × [ ( 6 ) 2 ] 4.5 ÷ [ ( 6 ) 3 ] 4.5 = ( 6 ) 6.5 × ( 6 ) 9 ÷ ( 6 ) 13.5 = ( 6 ) ( 6.5 + 9 - 13.5 ) = ( 6 ) 2 answer : option b"
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
b
multiply(power(const_60.0, 4.5), multiply(power(6, 6.5), power(36, 4.5)))
power(n0,n1)|power(n2,const_0.25)|power(n4,n5)|multiply(#0,#1)|multiply(#3,#2)|
general
a rectangular grass field is 75 m * 55 m , it has a path of 3.2 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ?
"area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 3.2 * 2 ) 2 * 2.5 = > 682 682 * 2 = rs . 1364 answer : d"
a ) s . 1350 , b ) s . 1327 , c ) s . 1328 , d ) s . 1364 , e ) s . 1927
d
multiply(subtract(rectangle_area(add(75, multiply(3.2, 2)), add(55, multiply(3.2, 2))), rectangle_area(75, 55)), 2)
multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)|
geometry
a fill pipe can fill 1 / 2 of cistern in 20 minutes . in how many minutes , it can fill 1 / 2 of the cistern ?
required time = 20 * 2 * 1 / 2 = 20 minutes answer is d
a ) 5 min , b ) 10 min , c ) 15 min , d ) 20 min , e ) 25 min
d
divide(20, 1)
divide(n2,n0)
physics
what is the average of 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 ?
"add 120 , 130 , 140 , 510 , 520 , 530 , 1115 , 1120 , and 1125 grouping numbers together may quicken the addition sum = 5310 5310 / 9 = 590 . d"
a ) 419 , b ) 551 , c ) 601 , d ) 590 , e ) 721
d
divide(add(add(add(add(add(add(1115, 130), 140), 510), 520), 530), 1115), add(const_3, const_4))
add(n6,n1)|add(const_3,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n6,#5)|divide(#6,#1)|
general
a bus travels from town a to town b . if the bus ' s speed is 50 km / hr , it will arrive in town b 42 min later than scheduled . if the bus increases its speed by 509509 m / sec , it will arrive in town b 30 min earlier than scheduled . find : a ) the distance between the two towns ; b ) the bus ' s scheduled time of arrival in b ; c ) the speed of the bus when it ' s on schedule .
first we will determine the speed of the bus following its increase . the speed is increased by 509509 m / sec = 50 ⋅ 60 ⋅ 6091000 = 50 ⋅ 60 ⋅ 6091000 km / hr = 20 = 20 km / hr . therefore , the new speed is v = 50 + 20 = 70 v = 50 + 20 = 70 km / hr . if xx is the number of hours according to the schedule , then at the speed of 50 km / hr the bus travels from a to b within ( x + 4260 ) ( x + 4260 ) hr . when the speed of the bus is v = 70 v = 70 km / hr , the travel time is x − 3060 x − 3060 hr . then 50 ( x + 4260 ) = 70 ( x − 3060 ) 50 ( x + 4260 ) = 70 ( x − 3060 ) 5 ( x + 710 ) = 7 ( x − 12 ) 5 ( x + 710 ) = 7 ( x − 12 ) 72 + 72 = 7 x − 5 x 72 + 72 = 7 x − 5 x 2 x = 72 x = 7 x = 72 x = 72 hr . so , the bus is scheduled to make the trip in 33 hr 3030 min . the distance between the two towns is 70 ( 72 − 12 ) = 70 ⋅ 3 = 21070 ( 72 − 12 ) = 70 ⋅ 3 = 210 km and the scheduled speed is 210 / 7 / 2 = 60 km / hr . answer : a
a ) 60 , b ) 37 , c ) 26 , d ) 28 , e ) 11
a
multiply(50, add(divide(subtract(multiply(divide(30, const_60), multiply(509509, const_3_6)), multiply(50, divide(42, const_60))), subtract(multiply(509509, const_3_6), 50)), divide(42, const_60)))
divide(n3,const_60)|divide(n1,const_60)|multiply(n2,const_3_6)|multiply(#0,#2)|multiply(n0,#1)|subtract(#2,n0)|subtract(#3,#4)|divide(#6,#5)|add(#7,#1)|multiply(n0,#8)
physics
how many numbers from 10 to 100000 are exactly divisible by 9 ?
"10 / 9 = 1 and 100000 / 9 = 11111 = = > 11111 - 1 = 11110 . answer : b"
a ) 900 , b ) 11110 , c ) 1100 , d ) 1200 , e ) 1400
b
add(divide(subtract(multiply(floor(divide(100000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1)
divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)|
general
if 10 litres of an oil of rs . 40 per litres be mixed with 5 litres of another oil of rs . 66 per litre then what is the rate of mixed oil per litre ?
"40 * 10 = 400 66 * 5 = 330 730 / 15 = 48.66 answer : a"
a ) rs . 48.66 , b ) rs . 51.03 , c ) rs . 54.17 , d ) rs . 55.33 , e ) none of the above
a
divide(add(multiply(10, 40), multiply(5, 66)), add(10, 5))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|
gain
tom wants to buy items costing $ 25.35 , $ 70.69 , and $ 85.96 . he earns $ 6.50 an hour doing odd jobs . if ten percent of his income is put aside for other purposes , how many hours must he work to earn the money he needs for his purchases ? round your answer to the nearest whole hour .
$ 6.50 x . 10 = $ . 65 is 10 % of his hourly income $ 6.50 - . 65 = $ 5.85 hourly amount available to spend $ 25.35 + $ 70.69 + $ 85.96 = $ 182 total needed $ 182 ÷ $ 5.85 = 31.11 . . . rounds to 31 hours correct answer c
a ) 8 hours , b ) 48 hours , c ) 31 hours , d ) 18 hours , e ) 28 hours
c
floor(divide(add(add(25.35, 70.69), 85.96), multiply(6.5, subtract(const_1, divide(const_1, const_10)))))
add(n0,n1)|divide(const_1,const_10)|add(n2,#0)|subtract(const_1,#1)|multiply(n3,#3)|divide(#2,#4)|floor(#5)
physics
a 600 meter long train crosses a signal post in 40 seconds . how long will it take to cross a 1.8 kilometer long bridge , at the same speed ?
s = 600 / 40 = 15 mps s = 1800 / 15 = 120 sec = 2 min . answer : b
a ) 4 min , b ) 2 min , c ) 8 min , d ) 9 min , e ) 3 min
b
divide(divide(multiply(1.8, const_1000), speed(600, 40)), const_60)
multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)
physics
a small college reduced its faculty by approximately 13 percent to 195 professors . what was the original number of faculty members ?
"f x is the original number of faculty members , then after 13 % reduction in faculty members number is . 87 x but we are given . 87 x = 195 x = 224 so the original number of faculty members is 224 correct answer - c"
a ) 182 , b ) 208 , c ) 224 , d ) 254 , e ) 302
c
divide(195, divide(subtract(const_100, 13), const_100))
subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)|
gain
the probability that a man will be alive for 10 more yrs is 3 / 4 & the probability that his wife will alive for 10 more yrs is 1 / 5 . the probability that none of them will be alive for 10 more yrs , is
"sol . required probability = pg . ) x p ( b ) = ( 1 — d x ( 1 — i ) = : x 1 = 1 / 5 ans . ( d )"
a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 1 / 5 , e ) 2
d
multiply(subtract(1, divide(3, 4)), subtract(3, divide(3, 5)))
divide(n4,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)|
general
a reduction of 20 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ?
"100 * ( 20 / 100 ) = 20 - - - 10 ? - - - 1 = > rs . 2 100 - - - 80 ? - - - 2 = > rs . 2.5 answer : c"
a ) 2.9 , b ) 2.2 , c ) 2.5 , d ) 2.1 , e ) 2.3
c
multiply(divide(divide(multiply(divide(20, const_100), 100), 10), multiply(divide(20, const_100), 100)), const_100)
divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)|
gain
a company that ships boxes to a total of 12 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter . )
"let ' s start with 4 minimum number of colors so single color code we can make 4 nos . now if we need to make 2 color combination out of 4 we can do so in 4 ! / 2 ! * 2 ! or 4 * 3 / 2 or 6 so total we can make 4 + 6 = 10 color combinations but we have 12 boxes so let ' s look at 5 we get 5 single color codes and out 5 color choices , we can choose 2 in 5 ! / 2 ! * 3 ! ways or 10 ways . so total we can have 5 + 10 = 15 color combinations . so , minimum number we need will be 5 ans is b . . . . . . ."
a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) 24
b
subtract(divide(factorial(subtract(divide(12, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(12, const_2), const_1))
divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)|
general
the present worth of a sum due sometime hence is rs . 576 and the banker ’ s gain is rs . 16 . the true discount is
solution t . d = √ p . w xb . g = √ 576 x 16 = 96 . answer d
a ) rs . 36 , b ) rs . 72 , c ) rs . 48 , d ) rs . 96 , e ) none
d
sqrt(multiply(576, 16))
multiply(n0,n1)|sqrt(#0)
gain
two trains are moving in the same direction at 90 kmph and 36 kmph . the faster train crosses a man in the slower train in 29 seconds . find the length of the faster train ?
"relative speed = ( 90 - 36 ) * 5 / 18 = 3 * 5 = 15 mps . distance covered in 29 sec = 29 * 15 = 435 m . the length of the faster train = 435 m . answer : b"
a ) 425 , b ) 435 , c ) 445 , d ) 455 , e ) 465
b
multiply(divide(subtract(90, 36), const_3_6), 29)
subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)|
physics
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 5 minutes , then the area of the park ( in sq . m ) is :
"perimeter = distance covered in 5 min . = ( 12000 / 60 ) x 5 m = 1000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1000 or x = 100 . length = 300 m and breadth = 200 m . area = ( 300 x 200 ) m 2 = 60000 m 2 . answer : d"
a ) 153601 , b ) 153600 , c ) 153602 , d ) 60000 , e ) 153604
d
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 5), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 5), const_1000), add(3, 2)), const_2), 2))
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
physics
the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs . 8748 , its purchase price was :
"explanation : = rs . 12000 answer : b ) 12000"
a ) 12003 , b ) 12000 , c ) 12002 , d ) 12289 , e ) 12019
b
divide(8748, subtract(const_1, multiply(divide(10, const_100), 3)))
divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2)|
gain
solve for x the equation log 9 ( x 3 ) = log 2 ( 8 )
log 9 ( x 3 ) = log 2 ( 8 ) : given log 2 ( 23 ) = 3 : simplify right hand side of given equation . log 9 ( x 3 ) = 3 : rewrite the above equation log 9 ( x 3 ) = log 9 ( 93 ) : rewite 3 as a log base 9 . x 3 = 93 : obtain algebraic equation from eqaution d . x = 9 : solve above for x correct answer a
a ) 9 , b ) 8 , c ) 7 , d ) 6 , e ) 5
a
power(const_2, divide(multiply(log(8), log(9)), 3))
log(n3)|log(n0)|multiply(#0,#1)|divide(#2,n1)|power(const_2,#3)
general
there are 6 more women than there are men on a local co - ed softball team . if there are a total of 16 players on the team , what is the ratio of men to women ?
"w = m + 6 w + m = 16 m + 6 + m = 16 2 m = 10 m = 5 w = 11 ratio : 5 : 11 ans : d"
a ) 10 / 16 , b ) 6 / 16 , c ) 4 / 16 , d ) 5 / 11 , e ) 4 / 10
d
divide(divide(subtract(16, 6), add(const_1, const_1)), add(divide(subtract(16, 6), add(const_1, const_1)), 6))
add(const_1,const_1)|subtract(n1,n0)|divide(#1,#0)|add(n0,#2)|divide(#2,#3)|
general
salad dressing p is made up of 30 % vinegar and 70 % oil , and salad dressing q contains 10 % vinegar and 90 % oil . if the two dressings are combined to produce a salad dressing that is 25 % vinegar , dressing p comprises what percentage of the new dressing ?
"let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.25 0.2 x = 0.15 x = 0.75 = 75 % the answer is a ."
a ) 75 % , b ) 60 % , c ) 40 % , d ) 25 % , e ) 20 %
a
divide(subtract(30, 10), subtract(25, 10))
subtract(n0,n2)|subtract(n4,n2)|divide(#0,#1)|
gain
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 135 grams of water ?
"( 16 / 18 ) * 144 = 120 grams the answer is a ."
a ) 120 , b ) 116 , c ) 112 , d ) 108 , e ) 104
a
multiply(2, divide(135, add(2, 16)))
add(n0,n1)|divide(n2,#0)|multiply(n0,#1)|
other
a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 63 km downstream .
"explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 63 / 21 = 3 hours option a"
a ) 3 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours
a
divide(63, add(16, 5))
add(n0,n1)|divide(n2,#0)|
physics
john and jacob set out together on bicycle traveling at 15 and 12 miles per hour , respectively . after 40 minutes , john stops to fix a flat tire . if it takes john two hour to fix the flat tire and jacob continues to ride during this time , how many hours will it take john to catch up to jacob assuming he resumes his ride at 15 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible )
"john ' s speed - 15 miles / hr jacob ' s speed - 12 miles / hr after 40 min ( i . e 2 / 3 hr ) , distance covered by john = 15 x 2 / 3 = 10 miles . jacob continues to ride for a total of 2 hour and 40 min ( until john ' s bike is repaired ) . distance covered in 2 hour 40 min ( i . e 8 / 3 hr ) = 12 x 8 / 3 = 32 miles . now , when john starts riding back , the distance between them is 22 miles . jacob and john are moving in the same direction . for john to catch jacob , the effective relative speed will be 15 - 12 = 3 miles / hr . thus , to cover 22 miles at 3 miles / hr , john will take 22 / 3 = 7 1 / 3 hours answer b"
a ) 7 , b ) 7 1 / 3 , c ) 7 1 / 2 , d ) 4 , e ) 7 1 / 2
b
divide(add(subtract(15, 12), subtract(12, subtract(multiply(divide(40, const_60), 15), multiply(divide(40, const_60), 12)))), subtract(multiply(divide(40, const_60), 15), multiply(divide(40, const_60), 12)))
divide(n2,const_60)|subtract(n0,n1)|multiply(n0,#0)|multiply(n1,#0)|subtract(#2,#3)|subtract(n1,#4)|add(#1,#5)|divide(#6,#4)|
physics
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 600 m long ?
"l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1800 m t = d / s t = 1800 / 10 t = 180 sec answer : b"
a ) 266 sec , b ) 180 sec , c ) 776 sec , d ) 166 sec , e ) 997 sec
b
divide(add(1200, 600), divide(1200, 120))
add(n0,n2)|divide(n0,n1)|divide(#0,#1)|
physics
two mba admissions committees are to be formed randomly from 6 second year mbas with 3 members each . what is the probability v that jane will be on the same committee as albert ?
"total number of ways to choose 3 member committee - 6 c 3 = ( 6 ! / 3 ! 3 ! ) = 20 no . of ways albert n jane are in same committee : - ( 4 c 1 * 2 ) = 8 probability v = ( 8 / 20 ) * 100 = 40 % . + 1 for me . . : d"
a ) 12 % , b ) 20 % , c ) 33 % , d ) 40 % , e ) 50 %
d
multiply(divide(multiply(choose(const_4, const_1), const_2), choose(6, 3)), multiply(multiply(const_5, const_5), const_4))
choose(const_4,const_1)|choose(n0,n1)|multiply(const_5,const_5)|multiply(#0,const_2)|multiply(#2,const_4)|divide(#3,#1)|multiply(#5,#4)|
probability
a rectangular plot 15 m × 10 m , has a path of grass outside it . if the area of grassy pathway is 54 m 2 , find the width of the path .
let the width of the path = w m then , length of plot with path = ( 15 + 2 w ) m and breadth of plot with path = ( 10 + 2 w ) m therefore , area of rectangular plot ( without path ) = 15 × 10 = 150 m 2 and area of rectangular plot ( with path ) = 150 + 54 = 204 m 2 hence , ( 15 + 2 w ) × ( 10 + 2 w ) = 204 ⇒ 4 w 2 + 50 w – 54 = 0 ⇒ 2 w 2 + 25 w – 27 = 0 ⇒ ( w – 2 ) ( w + 27 ) = 0 thus w = 2 or – 27 ∴ with of the path = 2 m answer c
['a ) 4 m', 'b ) 3 m', 'c ) 2 m', 'd ) 1 m', 'e ) none of these']
c
subtract(sqrt(add(54, power(divide(rectangle_perimeter(15, 10), const_4), const_2))), divide(rectangle_perimeter(15, 10), const_4))
rectangle_perimeter(n0,n1)|divide(#0,const_4)|power(#1,const_2)|add(n2,#2)|sqrt(#3)|subtract(#4,#1)
geometry
if a car had traveled 40 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 120 km , at what speed did it travel ?
"time = distance / speed difference in time = 1 / 2 hrs 120 / x - 120 / ( x + 40 ) = 1 / 2 substitute the value of x from the options . - - > x = 80 - - > 120 / 80 - 120 / 120 = 3 / 2 - 1 = 1 / 2 answer : d"
a ) 50 kmh , b ) 60 kmh , c ) 70 kmh , d ) 80 kmh , e ) 90 kmh
d
divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(120, 40)), const_4), power(40, const_2))), 40), const_2)
multiply(n0,n2)|power(n0,const_2)|multiply(#0,const_2)|multiply(#2,const_4)|add(#3,#1)|sqrt(#4)|subtract(#5,n0)|divide(#6,const_2)|
physics
a grocer has a sale of rs . 3435 , rs . 3927 , rs . 3855 , rs . 4230 and rs . 3562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 3500 ?
"explanation : total sale for 5 months = rs . ( 3435 + 3927 + 3855 + 4230 + 3562 ) = rs . 19009 . required sale = rs . [ ( 3500 x 6 ) â € “ 19009 ] = rs . ( 21000 â € “ 19009 ) = rs . 1991 . answer a"
a ) s . 1991 , b ) s . 2991 , c ) s . 3991 , d ) s . 4991 , e ) s . 5991
a
subtract(multiply(add(5, const_1), 3500), add(add(add(add(3435, 3927), 3855), 4230), 3562))
add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|
general
what is the probability of getting a number less than 4 when a die is rolled ?
"total number of outcomes possible when a die is rolled = 6 ( ∵ any one face out of the 6 faces ) i . e . , n ( s ) = 6 e = getting a number less than 4 = { 1 , 2 , 3 } hence , n ( e ) = 3 the probability = 3 / 6 = 1 / 2 . answer : b"
a ) 5 / 5 , b ) 1 / 2 , c ) 5 / 1 , d ) 5 / 7 , e ) 5 / 2
b
divide(const_2, choose(add(const_3, const_3), const_3))
add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)|
probability
a 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 second . what is the length of the other train ?
relative speed = 120 + 80 = 200 * 5 / 18 = 500 / 9 m / s let the length of the other train be x meters x + 270 / 9 = 500 / 9 x = 230 m answer is b
a ) 150 m , b ) 230 m , c ) 290 m , d ) 310 m , e ) 420 m
b
subtract(multiply(divide(add(120, 80), const_3_6), 9), 270)
add(n1,n2)|divide(#0,const_3_6)|multiply(n3,#1)|subtract(#2,n0)
physics
two trains of equal length , running with the speeds of 60 and 40 kmph , take 40 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ?
"rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 40 d = 40 * 100 / 18 = 2000 / 9 rs = 60 + 40 = 100 * 5 / 18 t = 2000 / 9 * 18 / 500 = 8 sec answer : e"
a ) 10 sec , b ) 16 sec , c ) 13 sec , d ) 67 sec , e ) 8 sec
e
multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 40), inverse(multiply(const_0_2778, add(60, 40))))
add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)|
physics
a dog takes 2 leaps for every 3 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is :
"explanation : dog : hare = ( 2 * 3 ) leaps of hare : 3 leaps of hare = 6 : 5 . answer : e ) 6 : 5"
a ) 1 : 5 , b ) 2 : 5 , c ) 3 : 5 , d ) 4 : 5 , e ) 6 : 5
e
divide(multiply(2, 3), 3)
multiply(n0,n2)|divide(#0,n1)|
other
in a group of pigs and hens , the number of legs are 22 more than twice the number of heads . the number of pigs is
explanation : let the number of pigs be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 22 4 x + 2 y = 2 x + 2 y + 22 2 x = 22 x = 11 answer : e
a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 11
e
divide(22, const_2)
divide(n0,const_2)
general
x + y = 19 , and x + 3 y = 1 . find the value of x + 2 y
add these two equations 2 x + 4 y = 20 divide by 2 ( to get x + 2 y ) answer will be d . 10
a ) 20 , b ) 18 , c ) 11 , d ) 10 , e ) 5
d
subtract(multiply(add(2, const_4), 2), 2)
add(n3,const_4)|multiply(n3,#0)|subtract(#1,n3)
general
in the fifth grade at parkway elementary school there are 470 students . 300 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ?
"total students = 470 boys = 300 , girls = 170 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 170 - 35 = 135 . correct option : b"
a ) 69 . , b ) 135 . , c ) 81 . , d ) 91 . , e ) 108 .
b
subtract(subtract(470, 300), subtract(250, divide(multiply(250, 86), const_100)))
multiply(n2,n3)|subtract(n0,n1)|divide(#0,const_100)|subtract(n2,#2)|subtract(#1,#3)|
gain