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for a certain alarm system , each code is comprised of 5 digits , but no digit can be used more than twice . how many codes can be made ?
case 1 : all digits distinct select 5 digits out of 10 in 10 c 5 ways and arrange them in 5 ! ways which gives us 10 c 5 x 5 ! = 30240 case 2 : two digits are same , other 3 are distinct e . g . 45722 select a digit which is to be repeated in 10 c 1 ways , select other 3 digits in 9 c 3 ways and arrange them in 5 ! / 2...
a ) 30240 , b ) 60480 , c ) 91440 , d ) 98240 , e ) 101040
c
add(add(divide(factorial(const_10), factorial(subtract(const_10, 5))), multiply(multiply(const_10, divide(factorial(subtract(const_10, const_1)), multiply(factorial(const_3), factorial(subtract(subtract(const_10, const_1), const_3))))), divide(factorial(5), const_2))), multiply(multiply(divide(factorial(const_10), mult...
factorial(const_10)|factorial(n0)|factorial(const_3)|factorial(const_2)|subtract(const_10,n0)|subtract(const_10,const_1)|subtract(const_10,const_2)|divide(#1,const_2)|divide(#1,const_4)|factorial(#4)|factorial(#5)|factorial(#6)|subtract(#5,const_3)|divide(#0,#9)|factorial(#12)|multiply(#11,#3)|divide(#0,#15)|multiply(#...
general
in the rectangular coordinate system , if the line x = 3 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ?
x = 3 y + 5 , and thus y = x / 3 - 5 / 3 the slope is 1 / 3 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) / ( m + 2 - m ) = p / 2 p / 2 = 1 / 3 and thus p = 2 / 3 the answer is e .
a ) - 2 , b ) 0 , c ) 1 / 2 , d ) 1 , e ) 2 / 3
e
divide(2, const_3)
divide(n2,const_3)
general
what decimal fraction is 30 ml of a litre ?
"answer required fraction = 30 / 1000 = 3 / 100 = . 03 correct option : d"
a ) . 3 , b ) . 0003 , c ) . 003 , d ) 0.03 , e ) none of these
d
divide(30, const_1000)
divide(n0,const_1000)|
physics
a can do a piece of work in 12 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in
"explanation : b ' s 5 days work = 1 / 10 βˆ— 5 = 1 / 2 remaining work = 1 βˆ’ 1 / 2 = 1 / 2 a can finish work = 12 βˆ— 1 / 2 = 6 days option b"
a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 9 days
b
divide(const_1, add(divide(const_1, 12), divide(const_1, 10)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|
physics
the average of 10 numbers was calculated as 20 . it is discovered later on that while calculating the average , one number , namely 86 , was incorrectly read as 26 . what is the correct average ?
"10 * 20 - 26 + 86 = 260 260 / 10 = 26 the answer is e ."
a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26
e
divide(add(multiply(10, 20), subtract(86, 26)), 10)
multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)|
general
find the compound interest on $ 1200 for 3 years at 20 % p . a . if ci is component yearly ?
"a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 3 = $ 2074 ci = $ 874 answer is c"
a ) $ 120 , b ) $ 150 , c ) $ 874 , d ) $ 250 , e ) $ 300
c
subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 3)), 1200)
divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)|
gain
the cross - section of a cannel is a trapezium in shape . if the cannel is 12 m wide at the top and 8 m wide at the bottom and the area of cross - section is 690 sq m , the depth of cannel is ?
"1 / 2 * d ( 12 + 8 ) = 690 d = 69 answer : d"
a ) 39 , b ) 28 , c ) 27 , d ) 69 , e ) 71
d
divide(divide(divide(690, divide(add(12, 8), const_2)), 8), const_2)
add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)|
physics
a cistern which could be filled in 7 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ?
1 / 7 - 1 / x = 1 / 10 = > 70 / 3 hrs answer : b
a ) 76 hrs , b ) 70 / 3 hrs , c ) 55 hrs , d ) 90 hrs , e ) 11 hrs
b
inverse(subtract(divide(const_1, 7), divide(const_1, const_10)))
divide(const_1,n0)|divide(const_1,const_10)|subtract(#0,#1)|inverse(#2)
physics
pooja travels from a to b a distance of 800 miles in 10 hours . he returns to a in 5 hours . find his average speed ?
speed from a to b = 800 / 8 = 100 mph speed from b to a = 800 / 5 = 160 mph average speed = 2 * 100 * 160 / 260 = 123.07 mph answer is a
a ) 123.07 mph , b ) 100 mph , c ) 150.07 mph , d ) 200 mph , e ) 300 mph
a
divide(add(divide(800, 10), divide(800, 5)), const_2)
divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)
physics
find the simple interest on rs . 68,000 at 16 2 / 3 % per year for 9 months .
"p = rs . 68000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . simple interest = ( p * r * t ) / 100 = rs . ( 68,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8500 answer is c ."
a ) 7500 , b ) 6500 , c ) 8500 , d ) 9500 , e ) none of them
c
multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100))
add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)|
gain
if y is the smallest positive integer such that 11,025 multiplied by y is the square of an integer , then y must be
"i just tried plugging in the numbers and found out that 4 * 11025 = 44 , 100 , which is a square of 210 b"
a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 14
b
multiply(add(const_2, const_3), const_2)
add(const_2,const_3)|multiply(#0,const_2)|
geometry
when jessica withdrew $ 400 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 4 of the remaining balance , what will be the final balance in her bank account ?
as per the question 400 = 2 a / 5 thus - a which is the total amount = 1000 the amount thus left = 600 she then deposited 1 / 4 of 600 = 150 total amount in her account = 750 answer e
a ) 300 , b ) 375 , c ) 400 , d ) 500 , e ) 750
e
multiply(subtract(divide(400, subtract(const_1, divide(const_3, 5))), 400), add(1, divide(1, 4)))
divide(n3,n4)|divide(const_3,n2)|add(n3,#0)|subtract(const_1,#1)|divide(n0,#3)|subtract(#4,n0)|multiply(#2,#5)
general
a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 290 m long ?
"explanation : speed = 45 km / hr = 45 Γ— ( 10 / 36 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 290 = 650 meter time taken to cross the platform = 650 / ( 25 / 2 ) = 650 Γ— 2 / 25 = 52 seconds answer : option e"
a ) 38 sec , b ) 35 sec , c ) 44 sec , d ) 40 sec , e ) 52 sec
e
multiply(divide(add(divide(290, const_1000), divide(360, const_1000)), 45), const_3600)
divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)|
physics
if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 81 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ?
"( x + 1 ) ^ 2 < = 81 x < = 8 x > = - 10 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 10 Γ— - 6 = 60 minimum possible value of xy is - 10 Γ— 8 = - 80 - 80 + 60 = - 20 answer : b"
a ) - 16 , b ) - 20 , c ) 0 , d ) 14 , e ) 16
b
add(sqrt(81), sqrt(64))
sqrt(n2)|sqrt(n5)|add(#0,#1)|
general
the simple interest on a sum of money will be rs . 900 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ?
"p - - - 10 - - - - 900 p - - - 5 - - - - - 450 3 p - - - 5 - - - - - 1350 - - - - - - = > 1800 answer : a"
a ) 1800 , b ) 2888 , c ) 1200 , d ) 2699 , e ) 2771
a
add(multiply(multiply(divide(900, 10), 5), const_3), multiply(divide(900, 10), 5))
divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)|
general
a number when divided by 6 leaves a remainder 3 . when the square of the same number is divided by 6 , the remainder is :
let x = 6 q + 3 . then , x 2 = ( 6 q + 3 ) 2 = 36 q 2 + 36 q + 9 = 6 ( 6 q 2 + 6 q + 1 ) + 3 . so , when 2 n is divided by 4 , remainder = 3 . answer : d
['a ) 0', 'b ) 1', 'c ) 2', 'd ) 3', 'e ) 4']
d
divide(multiply(6, 3), 6)
multiply(n0,n1)|divide(#0,n0)
geometry
the present ratio of students to teachers at a certain school is 40 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ?
"we are given that the ratio of students to teacher is 40 to 1 . we can rewrite this using variable multipliers . students : teachers = 40 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into...
a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15
a
divide(add(40, 25), 25)
add(n0,n4)|divide(#0,n4)|
other
what is the sum of all possible solutions to | x - 6 | ^ 2 + | x - 6 | = 20 ?
denote | x - 6 | as y : y ^ 2 + y = 20 - - > y = - 5 or y = 4 . discard the first solution since y = | x - 6 | , so it ' s an absolute value and thus can not be negative . y = | x - 6 | = 4 - - > x = 10 or x = - 2 . the sum = 12 . answer : d .
a ) - 1 , b ) 6 , c ) 7 , d ) 12 , e ) 14
d
add(add(const_4, 6), subtract(6, const_4))
add(n0,const_4)|subtract(n0,const_4)|add(#0,#1)
general
how many diagonals does a polygon with 15 sides have , if one of its vertices does not connect to any diagonal ?
if i calculate it using the formulae , # diagonals = n ( n - 3 ) / 2 each vertex sends of n - 3 diagonals n = 15 - 1 then 14 * ( 14 - 3 ) / 2 = 77 correct option : c
['a ) 80', 'b ) 90', 'c ) 77', 'd ) 88', 'e ) 99']
c
divide(multiply(subtract(15, const_1), subtract(subtract(15, const_1), const_3)), const_2)
subtract(n0,const_1)|subtract(#0,const_3)|multiply(#0,#1)|divide(#2,const_2)
geometry
a and b undertake to do a piece of work for rs . 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of a
"c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a ’ s share = rs . ( 600 * 4 / 8 ) = rs . 300 , b ' s share = rs . ( 600 * 3 / 8 ) = rs . 225 . c ' s share = rs . [ 600 - ( 300 + 225 Β» ) = rs . 75 . answer is a"
a ) 300 , b ) 450 , c ) 750 , d ) 800 , e ) none of them
a
multiply(divide(const_1, add(add(const_4, 3), const_1)), 600)
add(n3,const_4)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2)|
physics
a hall is 18 meters long and 9 meters wide . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , what is the volume of the hall ( in cubic meters ) ?
"2 hl + 2 hw = 2 lw h = lw / ( l + w ) volume = lwh = ( lw ) ^ 2 / ( l + w ) = 972 the answer is d ."
a ) 640 , b ) 758 , c ) 862 , d ) 972 , e ) 1024
d
volume_rectangular_prism(18, 9, divide(multiply(rectangle_area(18, 9), const_2), rectangle_perimeter(18, 9)))
rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|multiply(#0,const_2)|divide(#2,#1)|volume_rectangular_prism(n0,n1,#3)|
geometry
truck x is 13 miles ahead of truck y , which is traveling the same direction along the same route as truck x . if truck x is traveling at an average speed of 47 miles per hour and truck y is traveling at an average speed of 53 miles per hour , how long will it take truck y to overtake and drive 5 miles ahead of truck x...
relative speed = 53 - 47 = 6 miles per hour dist required = 13 + 5 = 18 miles time taken to overtake = 18 / 6 = 3 hours . e is the answer .
a ) 2 hours , b ) 2 hours 20 minutes , c ) 2 hours 30 minutes , d ) 2 hours 45 minutes , e ) 3 hours
e
divide(add(13, 5), subtract(53, 47))
add(n0,n3)|subtract(n2,n1)|divide(#0,#1)
physics
how many bricks , each measuring 125 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ?
"number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 125 x 11.25 x 6 ) = 1280 answer : a"
a ) 1280 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3400
a
divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(125, 11.25), 6))
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|
physics
the number of boxes in a warehouse can be divided evenly into 5 equal shipments by boat or 24 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ?
answer is the lcm of 5 and 24 = 120 answer e
a ) 27 , b ) 33 , c ) 54 , d ) 81 , e ) 120
e
multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 5)
multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_3)|multiply(n0,#2)
general
a sum of money amounts to rs . 6690 after 3 years and to rs . 10,035 after 6 years on compound interest . find the sum .
"let the sum be rs . p . then p ( 1 + r / 100 ) 3 = 6690 … ( i ) p ( 1 + r / 100 ) 6 = 10035 … ( ii ) on dividing , we get ( 1 + r / 100 ) 3 = 10025 / 6690 = 3 / 2 . substituting this value in ( i ) , we get : p * ( 3 / 2 ) = 6690 or p = ( 6690 * 2 / 3 ) = 4460 hence , the sum is rs . 4460 . answer : b"
a ) 4360 , b ) 4460 , c ) 4560 , d ) 4660 , e ) 4760
b
multiply(divide(subtract(10,035, 6690), subtract(multiply(6690, 6), multiply(3, 10,035))), const_100)
multiply(n0,n3)|multiply(n1,n2)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,const_100)|
gain
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 45 , what is the total amount ?
"x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 45 35 - - - ? = > 175 answer : c"
a ) 166 , b ) 105 , c ) 175 , d ) 177 , e ) 169
c
add(add(multiply(divide(const_100, 45), 45), multiply(divide(30, 45), 45)), 45)
divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)|
general
without stoppages , a train travels certain distance with an average speed of 100 km / h , and with stoppages , it covers the same distance with an average speed of 95 km / h . how many minutes per hour the train stops ?
due to stoppages , it covers 5 km less . time taken to cover 5 km = 5 Γ’  β€ž 100 h = 1 Γ’  β€ž 20 h = 1 Γ’  β€ž 20 Γ£ β€” 60 min = 3 min answer c
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
c
subtract(const_60, multiply(divide(95, 100), const_60))
divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)
physics
a tank is 25 m long , 12 m wide and 6 m deep . what is the cost of plastering its walls and bottom at the rate of 75 paise per sq . m ?
"total surface s = 2 lw + 2 lh + 2 wh = ( 25 * 12 ) + ( 2 * 25 * 6 ) + ( 2 * 12 * 6 ) = 300 + 300 + 144 = 744 m 2 therefore cost of plastering = 744 * 75 = 5800 paise = 558 answer a"
a ) 558 , b ) 502 , c ) 516 , d ) 612 , e ) 700
a
multiply(add(multiply(multiply(add(25, 12), const_2), 6), multiply(25, 12)), divide(75, const_100))
add(n0,n1)|divide(n3,const_100)|multiply(n0,n1)|multiply(#0,const_2)|multiply(n2,#3)|add(#4,#2)|multiply(#5,#1)|
physics
a man swims downstream 36 km and upstream 26 km taking 2 hours each time , what is the speed of the man in still water ?
"36 - - - 2 ds = 18 ? - - - - 1 26 - - - - 2 us = 13 ? - - - - 1 m = ? m = ( 18 + 13 ) / 2 = 15.5 answer : c"
a ) 9 , b ) 18.4 , c ) 15.5 , d ) 16.7 , e ) 13.4
c
divide(add(divide(26, 2), divide(36, 2)), const_2)
divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|
physics
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 19200 . a ' s share is ?
"x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 19200 = 6400 answer : c"
a ) 8876 , b ) 2765 , c ) 6400 , d ) 1298 , e ) 1261
c
multiply(multiply(const_1, const_12), divide(19200, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3))))
multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0)|
gain
a student chose a number , multiplied it by 3 , then subtracted 220 from the result and got 110 . what was the number he chose ?
"let x be the number he chose , then 3 β‹… x βˆ’ 220 = 110 3 x = 330 x = 110 correct answer c"
a ) 90 , b ) 100 , c ) 110 , d ) 120 , e ) 130
c
divide(add(110, 220), 3)
add(n1,n2)|divide(#0,n0)|
general
2 + 2 + 2 Β² + 2 Β³ . . . + 2 ^ 10
"2 + 2 = 2 ^ 2 2 ^ 2 + 2 ^ 2 = ( 2 ^ 2 ) * ( 1 + 1 ) = 2 ^ 3 2 ^ 3 + 2 ^ 3 = ( 2 ^ 3 ) * ( 1 + 1 ) = 2 ^ 4 so you can notice the pattern . . . in the end you will have 2 ^ 10 + 2 ^ 10 , which will give you 2 ^ 11 answer c"
a ) 2 ^ 9 , b ) 2 ^ 10 , c ) 2 ^ 11 , d ) 2 ^ 35 , e ) 2 ^ 37
c
divide(multiply(2, add(2, 2)), 2)
add(n0,n3)|multiply(n3,#0)|divide(#1,n1)|
general
a football field is 7200 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 3600 square yards ?
"answer a ) 9600 yards need 1200 lbs 1 yard will need 1200 / 7200 = 1 / 6 lbs 3600 yards will need 1 / 6 * 3600 yards = 600 lbs b"
a ) 450 , b ) 600 , c ) 750 , d ) 2400 , e ) 3200
b
multiply(3600, divide(1200, 7200))
divide(n1,n0)|multiply(n2,#0)|
geometry
{ ( 476 + 424 ) 2 - 4 x 476 x 424 } = ?
given exp . = [ ( a + b ) 2 - 4 ab ] , where a = 476 and b = 424 = [ ( 476 + 424 ) 2 - 4 x 476 x 424 ] = [ ( 900 ) 2 - 807296 ] = 810000 - 807296 = 2704 . correct option is c
a ) 2135 , b ) 6551 , c ) 2704 , d ) 6466 , e ) 1113
c
subtract(power(add(476, 424), 2), multiply(multiply(4, 476), 424))
add(n0,n1)|multiply(n0,n3)|multiply(n1,#1)|power(#0,n2)|subtract(#3,#2)
general
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 18 - meter platform and caught it after it had traveled 53.4 meters . how many times did the ball bounce ?
"ans : 6 division of total diatance travelled will be 18 + 18 + 9 + 4.5 + 2.25 + 1.125 + 0.5 ans b"
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
b
divide(divide(18, const_2), const_2)
divide(n0,const_2)|divide(#0,const_2)|
general
a certain telescope increases the visual range at a particular location from 90 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ?
"original visual range = 90 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 90 ) / 90 * 100 % = 2 / 3 * 100 % = 66.67 % answer e"
a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 66 2 / 3 %
e
multiply(divide(subtract(150, 90), 90), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
general
bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 35 more . what was his original selling price ?
"let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 35 0.07 = 35 p = 5...
a ) $ 180 , b ) $ 280 , c ) $ 420 , d ) $ 550 , e ) $ 700
d
multiply(divide(35, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100)))
divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|subtract(#5,#2)|divide(n3,#6)|multiply(#2,#7)|
general
a car left town a towards town b driving at a speed of v = 32 km / hr . after 3 hours on the road the driver stopped for 15 min in town c . because of a closed road he had to change his route , making the trip 28 km longer . he increased his speed to v = 40 km / hr but still he was 30 min late . find the distance the c...
the driver did not plan the stop at c . suppose it took x hours for him to get from c to b . then the distance is s = 40 β‹… x km . it took x βˆ’ 3060 βˆ’ 1560 = x βˆ’ 4560 = x βˆ’ 34 h to drive from c to b . the distance from c to b is 32 ( x βˆ’ 34 ) km , which is 28 km shorter than 40 β‹… x , i . e . 32 ( x βˆ’ 34 ) + 28 = 40 x 32 ...
a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118
c
add(multiply(const_3, 32), divide(40, const_2))
divide(n4,const_2)|multiply(n0,const_3)|add(#0,#1)
physics
total num of pupiles in 3 grades of scholl is 333 . the no of pupiles in grades 1 and 2 are un 3 : 5 ratio and 2 and 3 grades are 7 : 11 ratio . what is the strength of class that has highest no of pupils
ratio g 1 : g 2 = 3 : 5 and g 2 : g 3 = 7 : 11 so g 1 : g 2 : g 3 = 21 : 35 : 55 let the strength of three classes are 21 x , 35 x and 55 x respectively , then 21 x + 35 x + 55 x = 333 = > 111 x = 333 or x = 3 so strength of the class with highest number of pupils = 55 x = 55 * 3 = 165 answer : c
a ) 145 , b ) 155 , c ) 165 , d ) 175 , e ) 185
c
divide(333, add(add(multiply(divide(3, 5), divide(7, 11)), divide(7, 11)), const_1))
divide(n8,n9)|divide(n0,n5)|multiply(#1,#0)|add(#0,#2)|add(#3,const_1)|divide(n1,#4)
other
a train 150 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in :
relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 Γ— 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 150 Γ— 2 / 15 ) sec = 20 sec answer : a
a ) 20 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 40 sec
a
multiply(const_3600, divide(divide(150, const_1000), add(25, 2)))
add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)
physics
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 d 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
twenty four meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ?
"area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 24 = > x / 360 * 2 * pi * r + 2 r = 24 = > 2 a / r + 2 r = 24 = > a = r 12 - r ^ 2 = r 12 - r ^ 2 we will now max using derivations max value of a will found at a = 0 i . e 12 - 2 r = 0 r = 6 c"
a ) 2 √ 2 , b ) 2 √ 5 , c ) 6 , d ) 4 √ 2 , e ) none of these
c
divide(divide(multiply(add(const_10, const_4), const_2), const_2), const_2)
add(const_10,const_4)|multiply(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|
geometry
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 20 % profit ?
"let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 120 % of rs . 640 = 120 / 100 * 640 = rs . 768 . answer : e"
a ) 277 , b ) 960 , c ) 277 , d ) 266 , e ) 768
e
multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(20, const_100)))
divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)|
gain
how many positive integers , from 2 to 100 , inclusive , are not divisible by odd integers greater than 1 ?
the no will be of the form 2 ^ n to achieve this . since any other form will have odd no in the prime factorization . hence we need to find solution of n for equation - 1 < 2 ^ n < 101 . 2 ^ 6 = 64 , 2 ^ 7 - 128 hence n can take values from 1 to 6 . hence answer : b
a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 50
b
divide(subtract(multiply(multiply(2, multiply(2, multiply(2, multiply(2, 2)))), 2), const_4), const_10)
multiply(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,const_4)|divide(#5,const_10)
general
a and b finish the job in 15 days . while a , b and c can finish it in 6 days . c alone will finish the job in
"explanation : 6 = ( 15 * x ) / ( 15 + x ) 90 + 6 x = 15 x 9 x = 90 x = 10 answer : option a"
a ) 10 days , b ) 30 days , c ) 60 days , d ) 70 days , e ) 50 days
a
divide(multiply(6, 15), subtract(15, 6))
multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
physics
in x game of billiards , x can give y 10 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 100 ?
"x scores 60 while y score 50 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 50 = 60 . in x game of 100 points , y gives ( 100 - 60 ) = 40 points to c . d"
a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50
d
subtract(multiply(subtract(60, 10), const_3), multiply(subtract(60, 30), const_3))
subtract(n1,n0)|subtract(n1,n2)|multiply(#0,const_3)|multiply(#1,const_3)|subtract(#2,#3)|
general
find avrg speed if a man travels at speed of 16 km / hr up and 30 km / hr dawn at an altitude of 200 m .
"avg speed = 2 * x * y / ( x + y ) = 2 * 16 * 30 / ( 16 + 30 ) = 20.8 answer : c"
a ) 22.8 , b ) 21.8 , c ) 20.8 , d ) 28.8 , e ) 29.8
c
divide(multiply(multiply(const_2, 16), 30), add(16, 30))
add(n0,n1)|multiply(n0,const_2)|multiply(n1,#1)|divide(#2,#0)|
physics
a can do a work in 16 days and b can do it in 8 days . in how many days a and b can do the work ?
"explanation : a ' s 1 day ' s work = 1 / 16 b ' s 1 day ' s work = 1 / 8 they work together = 1 / 16 + 1 / 8 = 3 / 16 = 16 / 3 = 5 1 / 3 days answer : option b"
a ) 20 days , b ) 5 1 / 3 days , c ) 6 days , d ) 8 days , e ) 7 days
b
divide(const_1, subtract(divide(const_1, 8), divide(const_1, 16)))
divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|divide(const_1,#2)|
physics
in a forest 100 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged d...
"the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 100 tagged deers comprise 10 % of total # of deers - - > total # of deers = 100 * 10 = 1,000 . answer : d ."
a ) 150 , b ) 750 , c ) 1,250 , d ) 1,000 , e ) 2,500
d
multiply(50, 5)
multiply(n1,n3)|
general
find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 20 of money ?
"p = 20 ( 100 / 10 ) 2 = > p = 2000 answer : e"
a ) 1500 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2000
e
multiply(multiply(divide(20, multiply(10, 2)), const_100), multiply(10, 2))
multiply(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|multiply(#2,#0)|
general
a man sold 20 articles for $ 50 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 20 % ?
"production cost per article : $ 50 * ( 100 % - 20 % ) / 20 = $ 2.0 required production costs for a loss of 20 % : $ 90 * ( 100 % + 20 % ) = $ 108 number of articles to be sold for $ 108 to incur a 20 % loss : $ 108 / $ 2.0 = 54 thus , solution d is correct ."
a ) 45 , b ) 36 , c ) 40 , d ) 54 , e ) 48
d
divide(original_price_before_loss(20, 90), divide(original_price_before_gain(20, 50), 20))
original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2)|
gain
find the least number must be added to 433124 so that remaining no . is divisible by 17 ?
"on dividing 433124 by 17 we get the remainder 15 , so 2 should be added b"
a ) 3 , b ) 2 , c ) 5 , d ) 6 , e ) 1
b
subtract(17, reminder(433124, 17))
reminder(n0,n1)|subtract(n1,#0)|
general
a , b and c completed a piece of work , a worked for 6 days , b for 9 days and c for 4 days . their daily wages were in the ratio of 3 : 4 : 5 . find the daily wages of c , if their total earning was rs . 1554 ?
"3 x 4 x 5 x 6 9 4 18 x + 36 x + 20 x = 1554 74 x = 1554 = > x = 21 5 x = 105 rs . answer : e"
a ) s . 109 , b ) s . 108 , c ) s . 100 , d ) s . 103 , e ) s . 105
e
multiply(5, divide(1554, add(add(multiply(6, 3), multiply(9, 4)), multiply(4, 5))))
multiply(n0,n3)|multiply(n1,n2)|multiply(n2,n5)|add(#0,#1)|add(#3,#2)|divide(n6,#4)|multiply(n5,#5)|
physics
the average weight of 25 girls increases by 1 kg when a new girl comes in place of one of them weighing 55 kg . what might be the weight of the new girl ?
"total weight increased = 25 x 1 kg = 25 kg . weight of new person = 55 + 25 kg = 80 kg answer : d"
a ) 85 kg , b ) 90 kg , c ) 83 kg , d ) 80 kg , e ) 82 kg
d
add(multiply(25, 1), 55)
multiply(n0,n1)|add(n2,#0)|
general
everyone shakes hands with everyone else in a room . total number of handshakes is 120 . number of persons = ?
"in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 120 or n ( n - 1 ) = 240 or n = 16 answer is ( e )"
a ) a . 14 , b ) b . 12 , c ) c . 11 , d ) d . 15 , e ) e . 16
e
divide(divide(multiply(120, const_2), const_3), const_4)
multiply(n0,const_2)|divide(#0,const_3)|divide(#1,const_4)|
general
the average salary of the employees in a office is rs . 120 / month . the avg salary of officers is rs . 470 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office .
"let no . of non - officers be x 15 * 470 + x * 110 = ( x + 15 ) 120 x = 525 e"
a ) 400 , b ) 420 , c ) 430 , d ) 450 , e ) 525
e
divide(subtract(multiply(15, 470), multiply(15, 120)), subtract(120, 110))
multiply(n1,n3)|multiply(n0,n3)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)|
general
find the least number which when divided by 31 and 9 leaves a remainder of 3 in each case .
"the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 31 , 9 ) + 3 = 282 . answer : b"
a ) 280 , b ) 282 , c ) 284 , d ) 286 , e ) 288
b
add(3, lcm(31, 9))
lcm(n0,n1)|add(n2,#0)|
general
two train each 500 m long , are running in opposite directions on parallel tracks . if their speeds are 45 km / hr and 30 km / hr respectively , the time taken by the slower train to pass the driver of the faster one is ?
relative speed = 45 + 30 = 75 km / hr = 750 / 36 m / s = 125 / 6 m / s we are calculating the time taken by the slower train to pass the driver of the faster one . hence the distance = length of the smaller train = 500 m time = distance / speed = 500 / ( 125 / 6 ) = 24 s answer is b .
a ) 20 s , b ) 24 s , c ) 22 s , d ) 23 s , e ) 21 s
b
divide(500, multiply(add(45, 30), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)
physics
the average age of 32 students in a group is 10 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ?
"age of the teacher = ( 33 * 11 - 32 * 10 ) = 43 years . answer : b"
a ) 45 , b ) 43 , c ) 51 , d ) 53 , e ) 57
b
add(32, const_1)
add(n0,const_1)|
general
mohit sold an article for $ 21000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ?
"let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 21000 - 10 / 100 ( 21000 ) = 21000 - 2100 = $ 18900 = > 1.08 x = 18900 = > x = 17500 e"
a ) 16000 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17500
e
multiply(divide(subtract(21000, divide(multiply(21000, 10), const_100)), add(const_100, 8)), const_100)
add(n2,const_100)|multiply(n0,n1)|divide(#1,const_100)|subtract(n0,#2)|divide(#3,#0)|multiply(#4,const_100)|
gain
a marketing firm determined that , of 200 households surveyed , 80 used neither brand k nor brand b soap , 60 used only brand k soap , and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 households surveyed used both brands of soap ?
solution for soap k and soap b ( d ) 40
a ) 15 , b ) 20 , c ) 30 , d ) 40 , e ) 45
d
divide(subtract(subtract(200, 80), 60), add(3, const_1))
add(n3,const_1)|subtract(n0,n1)|subtract(#1,n2)|divide(#2,#0)
other
a shopkeeper cheats to the extent of 10 % while buying and selling , by using false weights . his total gain is .
answer : b ) 21 %
a ) 22 , b ) 21 , c ) 8 , d ) 9 , e ) 81
b
add(multiply(10, const_2), const_1)
multiply(n0,const_2)|add(#0,const_1)
gain
the number of singles that a baseball player hit increased 15 percent from her first season to her second season , and the number of doubles that she hit in the same time period decreased by 5 percent . if the total number of singles and doubles that she hit increased 4 percent from her first season to her second seaso...
soln : - 1.15 s + 0.95 d = 1.04 [ s + d ] 0.11 s = 0.09 d s / d = 9 / 11 answer : d
a ) 3 : 5 , b ) 5 : 7 , c ) 7 : 11 , d ) 9 : 11 , e ) 11 : 15
d
divide(subtract(const_100, 5), add(const_100, 15))
add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)
general
the product x of two prime numbers is between 17 and 55 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25 , then x =
"option b & c can be ruled out as they themselves are prime numbers 18 = 2 * 9 = 3 * 6 > > ignore 44 = 2 * 22 = 4 * 11 > > ignore 51 = 3 * 17 > > answer answer e"
a ) 18 , b ) 29 , c ) 37 , d ) 44 , e ) 51
e
multiply(add(2, const_1), subtract(25, 2))
add(n2,const_1)|subtract(n5,n2)|multiply(#0,#1)|
general
the average of first 6 prime numbers which are between 60 and 90 is
explanation : first six prime numbers which are between 60 and 90 = 61 , 67 , 71 , 73 , 79 , 83 average = ( 61 + 67 + 71 + 73 + 79 + 83 ) / 6 = 72.33 answer : e
a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 72.33
e
add(add(add(60, const_3), const_4), const_4)
add(n1,const_3)|add(#0,const_4)|add(#1,const_4)
general
a rectangular grass field is 95 m * 55 m , it has a path of 2.5 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 = ( 95 + 55 + 2.5 * 2 ) 2 * 2.5 = > 775 775 * 2 = rs . 1550 answer : b"
a ) 1350 , b ) 1550 , c ) 9676 , d ) 1679 , e ) 2691
b
multiply(subtract(rectangle_area(add(95, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(95, 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
the sale price sarees listed for rs . 350 after successive discount is 20 % and 5 % is ?
"350 * ( 80 / 100 ) * ( 95 / 100 ) = 266 answer : a"
a ) 266 , b ) 278 , c ) 342 , d ) 787 , e ) 191
a
subtract(subtract(350, divide(multiply(350, 20), const_100)), divide(multiply(subtract(350, divide(multiply(350, 20), const_100)), 5), const_100))
multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
a train is 360 meter long is running at a speed of 56 km / hour . in what time will it pass a bridge of 140 meter length .
"explanation : speed = 56 km / hr = 56 * ( 5 / 18 ) m / sec = 140 / 9 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 / 140 / 9 = 32.14 seconds answer : d"
a ) 20.14 seconds , b ) 12.14 seconds , c ) 22.14 seconds , d ) 32.14 seconds , e ) none of these
d
divide(add(360, 140), divide(multiply(56, const_1000), const_3600))
add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|
physics
a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 1100 , what is the reduced price for kg ?
"800 * ( 25 / 100 ) = 275 - - - - 5 ? - - - - 1 = > rs . 55 answer : c"
a ) s . 40 , b ) s . 46 , c ) s . 55 , d ) s . 41 , e ) s . 42
c
divide(divide(multiply(1100, 25), const_100), 5)
multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|
gain
in a college , 280 students study hindi , 254 students study sanskrit and 280 students study english . 97 students study hindi as well as physics , 138 students study sanskrit as well as english and 152 students study hindi as well as english . 73 students study all the 3 languages . how many total number of students a...
b
a ) 350 , b ) 500 , c ) 650 , d ) 235 , e ) 450
b
add(subtract(subtract(subtract(add(add(280, 254), 280), 97), 138), 152), 73)
add(n0,n1)|add(n0,#0)|subtract(#1,n3)|subtract(#2,n4)|subtract(#3,n5)|add(n6,#4)
general
20 men shake hands with each other . maximum no of handshakes without cyclic handshakes .
"1 st person will shake hand with 19 people 2 nd person will shake hand with 18 people 3 rd person will shake hand with 17 people . . . . . . total no . of handshakes = 19 + 18 + 17 + . . . + 3 + 2 + 1 = 19 * ( 19 + 1 ) / 2 = 190 or , if there are n persons then no . of shakehands = nc 2 = 20 c 2 = 190 answer : a"
a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 230
a
multiply(subtract(20, const_1), divide(20, const_2))
divide(n0,const_2)|subtract(n0,const_1)|multiply(#0,#1)|
general
a and b complete a work in 8 days . a alone can do it in 16 days . if both together can do the work in how many days ?
"1 / 8 + 1 / 16 = 3 / 16 16 / 3 = 5.3 days answer : c"
a ) 3.75 days , b ) 3.78 days , c ) 5.3 days , d ) 3.15 days , e ) 2.75 days
c
inverse(add(inverse(8), inverse(16)))
inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)|
physics
suppose 8 monkeys take 8 minutes to eat 8 bananas . how many monkeys would it take to eat 48 bananas in 48 minutes
"8 monkeys take 8 minutes to eat 8 bananas means each monkey takes 8 min for 1 banana in 48 mins one monkey can eat 6 bananas . . . . which means that 8 monkeys can eat 48 bananas in 48 minutes so answer is 8 monkeys answer : d"
a ) 5 monkeys , b ) 6 monkeys , c ) 7 monkeys , d ) 8 monkeys , e ) 9 monkeys
d
divide(48, divide(48, 8))
divide(n3,n0)|divide(n3,#0)|
physics
a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.2 per box per month for the record storage , what is the total amount that the company pays each month for record storage ?
volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.2 = 120 answer : d
a ) a . 150 , b ) b . 300 , c ) c . 600 , d ) d . 120 , e ) e . 200
d
multiply(0.2, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100)))
multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|divide(n3,#4)|multiply(n4,#5)
general
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 1.3 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 m...
"the easiest way for me is to plug in numbers . let the number of basic stereos produced be 40 , and number of delux stereos produced be 20 . total of 60 stereos . if it takes an hour to produce a basic stereo then it will take 1.3 hours to produce a deluxe stereo . 40 basic stereos = 40 hours . 20 delux stereos = 26 h...
a ) 7 / 17 , b ) 13 / 33 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2
b
divide(add(3, 2), const_10)
add(n0,n1)|divide(#0,const_10)|
general
two employees x and y are paid a total of rs . 528 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?
"let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 528 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 528 β‡’ y [ 12 / 10 + 1 ] = 528 β‡’ 22 y / 10 = 528 β‡’ 22 y = 5280 β‡’ y = 5280 / 22 = 480 / 2 = rs . 240 e"
a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 299 , e ) s . 240
e
divide(multiply(528, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2))
add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)|
general
8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 2 such machines to perform the same job ?
"let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 2 machines 2 units of work ( 2 machines ) - - - > 1 hour for 48 units of work 2 * 24 - - - > 1 * 24 hours d 24 hours"
a ) 2.25 hours , b ) 8.75 hours , c ) 12 hours , d ) 24 hours , e ) 16 hours
d
divide(multiply(6, 8), 2)
multiply(n0,n1)|divide(#0,n2)|
physics
in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 50 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ?
"working backward from the day it ' s covered : day 50 : fully covered day 49 : half covered so 49 days answer : d"
a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 49 , e ) 47
d
subtract(50, const_1)
subtract(n0,const_1)|
physics
a shopkeeper sold an article for rs 2564.36 . approximately what was his profit percent if the cost price of the article was rs 2200
"explanation : gain % = ( 364.36 * 100 / 2200 ) = 16.56 % = 17 % approx option a"
a ) 17 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 %
a
floor(multiply(const_100, divide(subtract(2564.36, 2200), 2200)))
subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|floor(#2)|
gain
if a ( a - 6 ) = 27 and b ( b - 6 ) = 27 , where a β‰  b , then a + b = ?
"i . e . if a = - 3 then b = 9 or if a = 9 then b = - 3 but in each case a + b = - 3 + 9 = 6 answer : option d"
a ) βˆ’ 48 , b ) βˆ’ 2 , c ) 46 , d ) 6 , e ) 48
d
subtract(subtract(subtract(subtract(add(add(6, 27), subtract(6, 27)), const_1), const_1), const_1), const_1)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)|
general
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 8.64 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discount rates is ...
let x be the discount on pony jeans . then 0.22 - x is the discount on fox jeans . 3 ( 0.22 - x ) ( 15 ) + 2 x ( 18 ) = 8.64 9.9 - 45 x + 36 x = 8.64 9 x = 1.26 x = 0.14 the answer is e .
a ) 8 % , b ) 10 % , c ) 11 % , d ) 12 % , e ) 14 %
e
multiply(subtract(divide(22, const_100), divide(subtract(8.64, multiply(divide(22, const_100), multiply(18, const_2))), subtract(multiply(15, const_3), multiply(18, const_2)))), const_100)
divide(n6,const_100)|multiply(n1,const_2)|multiply(n0,const_3)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)
gain
a cylindrical can has a radius of 4 centimeters and a height of 18 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ?
"they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 4 * 18 = 144 pi answer is d ."
a ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 144 pi , e ) 576 pi
d
multiply(multiply(4, 18), multiply(const_2, const_pi))
multiply(n0,n1)|multiply(const_2,const_pi)|multiply(#0,#1)|
geometry
what is 5 2 / 3 - 3 5 / 4 divided by 5 / 2 - 7 / 6 ?
"5 2 / 3 - 3 5 / 4 = 17 / 3 - 20 / 4 = ( 68 - 60 ) / 12 = 8 / 12 = 4 / 3 5 / 2 - 7 / 6 = ( 15 - 7 ) / 6 = 8 / 6 = 4 / 3 so 4 / 3 / 4 / 3 = 1 answer - e"
a ) 2 , b ) 6 , c ) 4 , d ) 5 , e ) 1
e
subtract(divide(add(multiply(const_10, 5), 5), 3), divide(add(const_10, 3), 4))
add(n3,const_10)|multiply(const_10,n0)|add(n0,#1)|divide(#0,n5)|divide(#2,n2)|subtract(#4,#3)|
general
the perimeter of a triangle is 36 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ?
"area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 36 / 2 = 45 cm 2 answer : c"
a ) 38 cm 2 , b ) 27 cm 2 , c ) 45 cm 2 , d ) 25 cm 2 , e ) 35 cm 2
c
triangle_area(2.5, 36)
triangle_area(n0,n1)|
geometry
sum of the squares of 3 no . ' s is 222 and the sum of their products taken two at a time is 131 . find the sum ?
"( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 222 + 2 * 131 a + b + c = √ 484 = 22 a"
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30
a
sqrt(add(multiply(131, const_2), 222))
multiply(n2,const_2)|add(n1,#0)|sqrt(#1)|
general
a salesman ' s income consists of a commission and a base salary of $ 450 per week . over the past 5 weeks , his weekly income totals have been $ 406 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 ov...
"total weekly income over 5 weeks = $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2070 = $ 1430 from this , we subtract base salary for 2 weeks i . e $ 450 * 2 = $ 900 therefore , commission = $ 1430 - $...
a ) $ 150 , b ) $ 245 , c ) $ 365 , d ) $ 715 , e ) $ 265
e
subtract(divide(subtract(multiply(500, 7), add(add(add(406, 413), add(436, 420)), 395)), const_2), 450)
add(n2,n3)|add(n4,n5)|multiply(n7,n8)|add(#0,#1)|add(n6,#3)|subtract(#2,#4)|divide(#5,const_2)|subtract(#6,n0)|
general
an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 400 sculptures are not on display , how many pieces of art does the gallery have ?
"too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 ...
a ) 360 , b ) 900 , c ) 540 , d ) 640 , e ) 720
b
divide(divide(400, subtract(1, divide(1, 3))), subtract(1, divide(1, 3)))
divide(n0,n1)|subtract(n0,#0)|divide(n6,#1)|divide(#2,#1)|
general
what is ( 25 ^ 7 + 25 ) / 25 ?
"( 25 ^ 7 + 25 ) / 25 = 25 * ( 25 ^ 6 + 1 ) / 25 = 25 ^ 6 + 1 clearly this is a number which ends with a 6 in the units place . the answer is a ."
a ) 244140626 , b ) 243120617 , c ) 242580531 , d ) 241230438 , e ) 240760355
a
divide(add(power(25, 7), 25), 25)
power(n0,n1)|add(n0,#0)|divide(#1,n0)|
general
chris mixed 5 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 3 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ?
"1 lbs of raisin = $ 1 5 lbs of raisin = $ 5 1 lbs of nuts = $ 3 4 lbs of nuts = $ 12 total value of mixture = 12 + 5 = 17 fraction of the value of raisin = 5 / 17 = 1 / 5 ans : a"
a ) 5 / 17 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 7
a
divide(5, add(multiply(4, 3), 5))
multiply(n1,n2)|add(n0,#0)|divide(n0,#1)|
general
a train running at the speed of 60 km / hr crosses a pole in 18 sec . what is the length of the train ?
"speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 18 = 300 m answer : c"
a ) 118 , b ) 150 , c ) 300 , d ) 258 , e ) 191
c
multiply(divide(multiply(60, const_1000), const_3600), 18)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
find the value for x from below equation : x / 3 = - 5 ?
"1 . multiply both sides by 3 : x * 3 / 3 = - 5 / 3 simplify both sides : x = - 15 c"
a ) - 66 , b ) 13 , c ) - 15 , d ) - 31 , e ) 40
c
multiply(subtract(const_2.0, const_4), 3)
subtract(const_2.0,const_4)|multiply(#0,n0)|
general
if an average hard drive had a capacity of 0.7 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ?
0.7 * 2 ^ 10 = 0.7 * 1024 = 716.8 the answer is e .
a ) 256 , b ) 512 , c ) 768 , d ) 1024 , e ) 7168
e
multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.7, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.7), const_10)
multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|multiply(#4,const_2)|multiply(#5,const_2)|multiply(#6,const_2)|multiply(#7,const_2)|multiply(#8,const_2)|add(n0,#9)|multiply(#10,const_10)
general
9 men went to a hotel . 8 of them spent rs . 3 each over their meals and the ninth spent rs . 2 more than the average expenditure of all the 9 . determine the total money spent by them ? a . rs . 29.25
average of 9 = x 9 x = 8 * 3 + x * 2 x = 3.25 total = 9 * 3.25 = 29.25 . answer : a
a ) 29.25 , b ) 29.28 , c ) 29.27 , d ) 29.22 , e ) 29.21
a
add(multiply(8, 3), divide(add(multiply(9, 2), multiply(8, 3)), subtract(9, const_1)))
multiply(n0,n3)|multiply(n1,n2)|subtract(n0,const_1)|add(#0,#1)|divide(#3,#2)|add(#4,#1)
general
a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 10 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 candidat...
"1 c 7 * 2 c 10 = 7 * 45 = 315 the answer is ( e )"
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 315
e
multiply(multiply(10, 3), 7)
multiply(n3,n5)|multiply(n1,#0)|
other
find the average of all the numbers between 6 and 34 which are divisible by 5
"exp . multiples of 5 between 6 and 34 are 10 , 15,20 , 25,30 average = ( 10 + 15 + 20 + 25 + 30 ) / 5 = 5 ( 10 + 30 ) / 2 * 5 = 40 / 2 = 20 answer : b"
a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 40
b
divide(add(add(6, const_4), subtract(34, const_4)), const_2)
add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|
general
how much time will a train of length 420 m moving at a speed of 72 kmph take to cross another train of length 640 m , moving at 36 kmph in the same direction ?
"the distance to be covered = sum of their lengths = 420 + 640 = 1060 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 1060 / 10 = 106 sec . answer : e"
a ) 50 , b ) 99 , c ) 88 , d ) 76 , e ) 106
e
divide(add(420, 640), multiply(subtract(72, 36), const_0_2778))
add(n0,n2)|subtract(n1,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
machine p and machine q are each used to manufacture 110 sprockets . it takes machine p 10 hours longer to produce 110 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?
p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 10 / x = 110 / 1.1 x + 10 1.1 ( 110 ) = 110 + 11 x 11 x = 11 x = 1 the answer is e
a ) 5 , b ) 15 , c ) 55 , d ) 95 , e ) 1
e
divide(subtract(110, divide(110, add(divide(10, const_100), const_1))), 10)
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)
gain
a spider web consists of 3 concurrent lines forming radial support for 20 concentric regular hexagon . the hexagon vertices are 5 mm from their immidiate neighbour . if the innermost hexagon has a side length 3 mm how many meters is the length of the entire web including the 3 diagonals ?
for perimeter of all hexagon = 6 * ( 3 + 8 + 13 + . . . . . . . . . . . 20 th term . ) sum will be 6060 mm and for length of one diagonal find 20 th term = 98 so total length = 6060 + 294 = 6354 mm = 6.354 m answer : c
['a ) 7.354 m', 'b ) 8.354 m', 'c ) 6.354 m', 'd ) 9.354 m', 'e ) 5.354 m']
c
add(multiply(multiply(3, const_2), add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(add(divide(3, const_1000), add(divide(3, const_1000), divide(5, const_1000))), add(divide(5, const_1000), add(divide(3, const_1000), divide(5, const_1000)))), add(divide(5, const_1000), add(divide(5, const_1000), ...
divide(n0,const_1000)|divide(n2,const_1000)|multiply(n0,const_2)|add(#0,#1)|add(#3,#0)|add(#3,#1)|add(#4,#5)|add(#5,#1)|add(#6,#7)|add(#7,#1)|add(#8,#9)|add(#9,#1)|add(#10,#11)|add(#11,#1)|add(#12,#13)|add(#13,#1)|add(#14,#15)|add(#15,#1)|add(#16,#17)|add(#17,#1)|add(#18,#19)|add(#19,#1)|add(#20,#21)|add(#21,#1)|add(#2...
physics
when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 25 % water ?
"explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 75 % ( 100 - 25 ) . therefore 0.5 x amount of flower - nectar = 0.75 x amount of honey = 0.75 x 1 kg therefore amount of flower - nectar needed = ( 0.75 / 0.51 ) kg = 1.5 kgs answer : b"
a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these
b
divide(divide(subtract(const_100, 25), const_100), divide(50, const_100))
divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0)|
general
a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 30 % impurities to the product . find the net profit percentage of the dealer ?
"the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 30 % impurities to the product . it means that he is selling $ 5 worth of product for $ 5.5 . so his profit is $ 5.5 - $ 4...
a ) 44 % , b ) 40 % , c ) 50 % , d ) 56.25 % , e ) 37.5 %
e
multiply(subtract(divide(add(const_100, 30), subtract(const_100, 20)), const_1), const_100)
add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100)|
general