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if k ^ 3 is divisible by 168 , what is the least possible value of integer k ?
k ^ 3 = 168 * x = 2 ^ 3 * 3 * 7 * x the factors of k must at minimum include 2 * 3 * 7 = 42 . the answer is b .
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60
b
divide(divide(168, const_2), const_2)
divide(n1,const_2)|divide(#0,const_2)
general
a and b can do a work in 20 days , b and c in 25 days , c and a in 30 days . if a , b and c work together , they will complete the work in ?
"a + b 1 day work = 1 / 20 b + c 1 day work = 1 / 25 c + a 1 day work = 1 / 30 adding we get 2 ( a + b + c ) = 1 / 20 + 1 / 25 + 1 / 30 = 37 / 300 a + b + c 1 day work = 37 / 600 a , b , c can finish the work in 600 / 37 days = 16 days approximately answer is c"
a ) 10 days , b ) 12 days , c ) 16 days , d ) 20 days , e ) 25 days
c
divide(const_1, divide(add(add(inverse(20), inverse(25)), inverse(30)), const_2))
inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|divide(const_1,#5)|
physics
in what time will a train 100 m long cross an electric pole , it its speed be 136 km / hr ?
"speed = 136 * 5 / 18 = 37.8 m / sec time taken = 100 / 37.8 = 2.6 sec . answer : a"
a ) 2.6 sec , b ) 1.9 sec , c ) 8.9 sec , d ) 6.9 sec , e ) 2.9 sec
a
divide(100, multiply(136, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
5220 / ? * 15 = 2175
"b ( 5220 / ? ) * 15 = 2175 ? = ( 5220 * 15 ) / 2175 = > ? = 36"
a ) 34 , b ) 36 , c ) 33 , d ) 42 , e ) 24
b
divide(5220, 15)
divide(n0,n1)|
general
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department take...
answer : d 126 ° divided by 360 ° equals 0.35 , therefore the sector is equal to 35 % of the total
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 70 %
d
multiply(divide(126, divide(const_3600, const_10)), const_100)
divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)|
physics
what is the unit digit in 14 ^ 100 ?
unit digit in 14 ^ 100 = unit digit in [ ( 14 ^ 4 ) ^ 25 ] but unit digit in 14 ^ 4 = 6 unit digit 14 ^ 100 = 6 answer is e
a ) 0 , b ) 5 , c ) 2 , d ) 3 , e ) 6
e
reminder(power(14, const_2), const_10)
power(n0,const_2)|reminder(#0,const_10)|
general
if n is the smallest integer such that 864 times n is the square of an integer , what is the value of n ?
"865 can written as = 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 - - > 2 ^ 5 * 3 ^ 3 - - - ( 1 ) so for 864 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 3 to make the power of 3 in ( 1 ) even option b is correct"
a ) 2 , b ) 3 , c ) 6 , d ) 12 , e ) 24
b
divide(divide(divide(divide(divide(divide(864, const_2), const_2), const_2), const_2), const_3), const_3)
divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_3)|divide(#4,const_3)|
geometry
tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 5 even numbers . his mother immediately worked out the answer . how much had he scored in math ?
"b 30 sum = ( n x n ) + n hence , 5 x 5 = 25 + 5 = 30"
a ) 80 , b ) 30 , c ) 40 , d ) 70 , e ) 60
b
multiply(add(5, const_1), 5)
add(n0,const_1)|multiply(n0,#0)|
physics
what is the sum of all digits for the number 10 ^ 25 - 58 ?
"10 ^ 25 is a 26 - digit number : 1 followed by 25 zeros . 10 ^ 25 - 58 is a 25 - digit number : 23 9 ' s and 42 at the end . the sum of the digits is 23 * 9 + 4 + 2 = 213 . the answer is e ."
a ) 197 , b ) 201 , c ) 205 , d ) 209 , e ) 213
e
multiply(add(divide(subtract(subtract(25, 10), const_2), const_2), 10), divide(add(subtract(25, 10), const_2), const_2))
subtract(n1,n0)|add(#0,const_2)|subtract(#0,const_2)|divide(#2,const_2)|divide(#1,const_2)|add(n0,#3)|multiply(#5,#4)|
general
bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 24 white marbles . how many red marbles could be in bag a ?
"6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 24 = 6 k + 4 p = > k = 2 , p = 3 total red ' s in bag a will be 2 k = 4 c"
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 8
c
divide(24, add(multiply(3, 2), 4))
multiply(n1,n2)|add(n5,#0)|divide(n6,#1)|
other
a can do a piece of work in 40 days ; b can do the same in 30 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ?
"10 / 40 + 10 / 30 + 10 / x = 1 x = 24 days answer : a"
a ) 24 days , b ) 65 days , c ) 86 days , d ) 45 days , e ) 17 days
a
divide(10, subtract(const_1, divide(add(10, 10), 40)))
add(n2,n2)|divide(#0,n1)|subtract(const_1,#1)|divide(n2,#2)|
physics
what least value should be replaced by * in 2631 * 4 so the number become divisible by 3
"explanation : trick : number is divisible by 3 , if sum of all digits is divisible by 3 , so ( 2 + 6 + 3 + 1 + * + 4 ) = 16 + * should be divisible by 9 , 16 + 2 will be divisible by 2 , so that least number is 2 . answer : option d"
a ) 3 , b ) 4 , c ) 5 , d ) 2 , e ) 7
d
subtract(4, subtract(4, 4))
subtract(n1,n1)|subtract(n1,#0)|
general
the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 735 , krishan has ?
ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 735 : n & there n = 289 x 735 / 49 = rs . 4335 answer : a
a ) s . 4335 , b ) s . 2330 , c ) s . 1190 , d ) s . 1620 , e ) s . 2680
a
multiply(divide(multiply(divide(735, 7), 17), 7), 17)
divide(n4,n0)|multiply(n1,#0)|divide(#1,n0)|multiply(n1,#2)
other
a train 146 m long is running at a speed of 98 kmph . how long does it take to pass a man who is running at 12 kmph in the same direction as the train ?
"answer : a . speed of the train relative to man = ( 98 - 12 ) kmph = ( 86 * 5 / 18 ) m / sec = ( 215 / 9 ) m / sec time taken by the train to cross the man = time taken by it to cover 146 m at 219 / 9 m / sec = 146 * 9 / 219 sec = 6 sec"
a ) 6 sec , b ) 9 sec , c ) 12 sec , d ) 18 sec , e ) 15 sec
a
divide(146, multiply(subtract(98, 12), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 5 screws and 1 nails ?
let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 5 s and 1 n = 5 s + 1 * 5 s = 10 s . . ratio = 15 s / 10 s = 15 / 10 = 3 / 2 d
a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 5 / 4
d
divide(add(multiply(3, const_4), 3), const_10)
multiply(n1,const_4)|add(n1,#0)|divide(#1,const_10)
general
a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 5 , and the ratio of soap to salt is four times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is quartered by water evaporation . at th...
"water : soap = 3 : 5 soap : salt = 12 : 20 = > for 12 soap , salt = 20 = > for 5 soap , salt = ( 20 / 12 ) * 5 = 100 / 12 = 25 / 3 so , water : soap : salt = 3 : 5 : 25 / 3 = 9 : 15 : 25 after open container , water : soap : salt = 2.25 : 15 : 25 so , water : salt = 2.25 : 25 = 9 : 100 answer : a"
a ) 9 : 100 , b ) 100 : 9 , c ) 12 : 100 , d ) 12 : 20 , e ) 13 : 100
a
divide(multiply(multiply(5, 3), 3), multiply(multiply(5, 5), 5))
multiply(n0,n1)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)|
other
the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 94 per day . what was his wage on the 8 th day ?
"average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 94 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 94 * 7 ) = 1350 - ( 609 + 658 ) = 1350 - 1267 = 83 answer a"
a ) $ 83 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 104
a
subtract(multiply(90, 15), add(multiply(87, 7), multiply(94, 7)))
multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|
physics
at what rate percent on simple interest will rs . 750 amount to rs . 1200 in 5 years ?
"450 = ( 750 * 5 * r ) / 100 r = 12 % . answer : a"
a ) 12 , b ) 3 , c ) 14 , d ) 5 , e ) 6
a
multiply(divide(divide(subtract(1200, 750), 750), 5), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
the difference between simple and compound interest on rs . 1200 for one year at 10 % per annum reckoned half - yearly is ?
"s . i . = ( 1200 * 10 * 1 ) / 100 = rs . 120 c . i . = [ 1200 * ( 1 + 5 / 100 ) 2 - 1200 ] = rs . 123 difference = ( 123 - 120 ) = rs . 3 . answer : b"
a ) rs . 7 , b ) rs . 3 , c ) rs . 8 , d ) rs . 1 , e ) rs . 4
b
multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1200)
divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6)|
gain
stacy and heather are 25 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the tw...
ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 25 - 2.4 = 22.6 time taken = distance / speed = 22.6 / 11 = 2.05 hrs heather wi...
a ) 9.0 mile , b ) 13.0 mile , c ) 11.0 mile , d ) 10.2 mile , e ) 12 mile
d
multiply(divide(subtract(25, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5)
add(n1,n2)|add(n2,#0)|divide(#0,const_60)|multiply(n3,#2)|subtract(n0,#3)|divide(#4,#1)|multiply(n2,#5)
physics
ganesh , ram and sohan together can do a work in 16 days . if ganesh and ram together can do a the same work in 24 days then , how long will take sohan to do the same work ?
solution : work done by three of them together in 1 day = 1 / 16 . . . . . . . . . . . . ( 1 ) work done by ganesh and ram together in 1 day = 1 / 24 . . . . . ( 2 ) so , work done by sohan in one day = ( 1 ) - ( 2 ) = ( 1 / 16 ) - ( 1 / 24 ) = 1 / 48 = > total days taken by sohan to complete the work alone = 48 days ....
a ) 48 , b ) 58 , c ) 68 , d ) 78 , e ) 100
a
divide(const_1, subtract(divide(const_1, 16), divide(const_1, 24)))
divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)
physics
one fourth of one third of two fifth of a number is 25 . what will be 40 % of that number
"explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 25 then x = 25 * 30 = 750 40 % of 750 = 300 answer : option e"
a ) a ) 140 , b ) b ) 150 , c ) c ) 180 , d ) d ) 200 , e ) e ) 300
e
divide(multiply(divide(25, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100)
add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)|
gain
the cost price of an article is 56 % of the marked price . calculate the gain percent after allowing a discount of 13 % .
"sol . let marked price = rs . 100 . then , c . p . = rs . 56 . s . p = rs . 87 . â ˆ ´ gain % = [ 31 / 56 * 100 ] % = 55.3 % . answer c"
a ) 23.4 % , b ) 44.5 % , c ) 55.3 % , d ) 57.4 % , e ) none
c
multiply(subtract(divide(subtract(const_100, 13), 56), const_1), const_100)
subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)|
gain
two trains each 260 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively . in what time will they cross each other completely ?
"explanation : d = 260 m + 260 m = 520 m rs = 80 + 70 = 150 * 5 / 18 = 125 / 3 t = 520 * 3 / 125 = 12.48 sec answer : option c"
a ) 15 sec , b ) 19 sec , c ) 12.48 sec , d ) 10 sec , e ) 11 sec
c
divide(260, multiply(80, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
1024 people in village 720 speak tamil , 562 speak english while 346 speak both . person who don not speack english or tamil speak hindi . what is probibility that person speak hindi
n ( t ) = 720 n ( e ) = 562 n ( t ^ e ) = 346 n ( tue ) = 720 + 562 - 346 = 936 now , n ( h ) = 1024 - 936 = 88 p ( h ) = 88 / 1024 = 11 / 128 = 0.085 answer : e
a ) 13 / 128 , b ) 12 / 128 , c ) 11 / 126 , d ) 11 / 127 , e ) 11 / 128
e
divide(subtract(1024, add(add(subtract(562, 346), subtract(720, 346)), 346)), 1024)
subtract(n2,n3)|subtract(n1,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)|divide(#4,n0)
other
a train 140 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is
"speed of train relative to man : 140 / 6 * 18 / 5 km / hr = 84 km / hr let speed of train = x therefore x + 6 = 84 x = 84 - 6 x = 78 km / hr answer : e"
a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 78 kmph
e
divide(divide(subtract(140, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778)
multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
the speed of a train is 120 kmph . what is the distance covered by it in 8 minutes ?
"120 * 8 / 60 = 16 kmph answer : c"
a ) 15 kmph , b ) 11 kmph , c ) 16 kmph , d ) 18 kmph , e ) 12 kmph
c
multiply(divide(8, const_60), 120)
divide(n1,const_60)|multiply(n0,#0)|
physics
a goods train runs at the speed of 72 kmph and crosses a 260 m long platform in 26 seconds . what is the length of the goods train ?
"speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 260 / 26 = 20 x + 260 = 520 x = 260 . answer : a"
a ) 260 m , b ) 270 m , c ) 643 m , d ) 832 m , e ) 270 m
a
multiply(subtract(26, divide(260, multiply(const_0_2778, 72))), multiply(const_0_2778, 72))
multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2)|
physics
if albert ’ s monthly earnings rise by 14 % , he would earn $ 678 . if , instead , his earnings rise by only 15 % , how much ( in $ ) would he earn this month ?
"= 678 / 1.14 ∗ 1.15 = 683 = 683 answer is c"
a ) 643 , b ) 689 , c ) 683 , d ) 690 , e ) 693
c
multiply(divide(678, add(const_1, divide(14, const_100))), add(const_1, divide(15, const_100)))
divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)|
gain
a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 45 minutes . the total distance is ?
"b 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 45 / 60 47 x / 60 = 45 / 60 = > x = 0.95 . total distance = 3 * 0.95 = 2.87 km ."
a ) 1 km , b ) 2.87 km , c ) 3 km , d ) 4 km , e ) 5 km
b
multiply(multiply(divide(divide(45, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000)
divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)|
physics
a container contains 40 litres of milk . from this container 4 litres of milk was taken out and replaced by water . this process was repeated further two times . how much milk is now contained by the container .
solution amount of milk left after 3 operations [ 40 ( 1 - 4 / 40 ) 3 ] litres = ( 40 x 9 / 10 x 9 / 10 x 9 / 10 ) = 29.16 litres answer d
a ) 26.34 litres , b ) 27.36 liters , c ) 28 litres , d ) 29.16 litres , e ) can not be determined
d
multiply(power(subtract(const_1, divide(4, 40)), const_3), 40)
divide(n1,n0)|subtract(const_1,#0)|power(#1,const_3)|multiply(n0,#2)
general
if grapes are 90 % water and raisins are 17 % water , then how many kilograms did a quantity of raisins , which currently weighs 14 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transfor...
"let x be the original weight of the grapes . the weight of the grape pulp was 0.1 x . since the grape pulp is 83 % of the raisins , 0.1 x = 0.83 ( 14 kg ) . then x = 8.3 * 14 = 116.2 kg . the answer is e ."
a ) 108.6 , b ) 110.4 , c ) 112.6 , d ) 114.8 , e ) 116.2
e
divide(multiply(divide(subtract(const_100, 17), const_100), 14), divide(subtract(const_100, 90), const_100))
subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)|
general
on multiplying a number f by 153 , the result obtained was 102325 . however , it is found that both the 2 ' s are wrong . find the correct result .
the only thing you actually know about the correct number f is that it is divisible by 153 and has 5 as a factor . you should immediately try to find the factors of 153 and look for them in the options . 153 = 9 * 17 divisibility by 9 is easy to check . only ( d ) satisfies .
a ) 104345 , b ) 107375 , c ) 108385 , d ) 109395 , e ) 105355
d
add(add(multiply(add(const_3, const_4), const_10), multiply(multiply(add(const_3, const_4), const_100), const_10)), 102325)
add(const_3,const_4)|multiply(#0,const_10)|multiply(#0,const_100)|multiply(#2,const_10)|add(#1,#3)|add(n1,#4)
other
at an election 2 candidates are participated a candidate got 30 % of votes and defeated by 5000 votes . 100 votes are invalid . find the total polled votes ?
winners votes = 100 - 30 = 70 % invalid votes = 100 polled votes = [ ( 100 * 5000 ) / ( 2 * 70 - 100 ) ] + 100 = 12600 answer is b
a ) 11520 , b ) 12600 , c ) 15000 , d ) 13560 , e ) 12560
b
add(const_100, multiply(divide(5000, subtract(subtract(100, 30), 30)), const_100))
subtract(n3,n1)|subtract(#0,n1)|divide(n2,#1)|multiply(#2,const_100)|add(#3,const_100)
gain
the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ?
"4 : 5 3 : 2 - - - - - - - 12 : 15 : 10 12 : 10 6 : 5 answer : c"
a ) 6 : 8 , b ) 6 : 2 , c ) 6 : 5 , d ) 6 : 4 , e ) 6 : 2
c
divide(multiply(4, 3), multiply(5, 2))
multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)|
other
if a large pizza has a radius that is 10 % larger that that of a medium pizza , what is the percent increase in area between a medium and a large pizza ?
"let the radius of medium pizza be r . then the radius of large pizza is 1.1 r . the area of the medium pizza is pi * r ^ 2 the area of the large pizza is pi * ( 1.1 * r ) ^ 2 = 1.21 * pi * r ^ 2 , an increase of 21 % . the answer is b ."
a ) 15 % , b ) 21 % , c ) 25 % , d ) 30 % , e ) 36 %
b
multiply(subtract(power(add(const_1, divide(10, const_100)), const_2), const_1), const_100)
divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|
gain
if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 2 flips and not on the last 3 flips ?
( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b
a ) 3 / 5 , b ) 1 / 32 , c ) 1 / 5 , d ) 1 / 4 , e ) 1 / 2
b
inverse(power(2, 5))
power(n1,n2)|inverse(#0)|
probability
60 kg of an alloy a is mixed with 100 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 2 and alloy b has tin and copper in the ratio 1 : 4 , then the amount of tin in the new alloy is ?
quantity of tin in 60 kg of a = 60 * 2 / 5 = 24 kg quantity of tin in 100 kg of b = 100 * 1 / 5 = 20 kg quantity of tin in the new alloy = 24 + 20 = 44 kg answer is b
a ) 24 kg , b ) 44 kg , c ) 20 kg , d ) 30 kg , e ) 52 kg
b
add(multiply(divide(60, add(3, 2)), 2), multiply(divide(100, add(1, 4)), 1))
add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)
other
what is the sum of all possible solutions to the equation √ 2 x ^ 2 − x − 9 = x + 1 ?
"first of all notice that since lhs is the square root of a number , it must be non - negative ( the square root function can not give negative result ) , then the rhs must also be non - negative : x + 1 ≥ 0 - - > x ≥ − 1 square the equation : 2 x ^ 2 − x − 9 = x ^ 2 + 2 x + 1 - - > x ^ 2 − 3 x − 10 = 0 - - > x = - 2 o...
a ) - 2 , b ) 2 , c ) 3 , d ) 5 , e ) 6
d
add(add(const_4, 2), subtract(9, const_4))
add(n0,const_4)|subtract(n2,const_4)|add(#0,#1)|
general
pipe a can fill a tank in 10 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ?
"let the pipe b be closed after x minutes . 30 / 10 - x / 24 = 1 = > x / 24 = 30 / 10 - 1 = 2 = > x = 2 * 24 = 48 . answer : e"
a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 48
e
multiply(subtract(divide(30, 10), const_1), 24)
divide(n2,n0)|subtract(#0,const_1)|multiply(n1,#1)|
physics
if price of t . v set is reduced by 18 % , then its sale increases by 88 % , find net effect on sale value
"- a + b + ( ( - a ) ( b ) / 100 ) = - 18 + 88 + ( - 18 * 88 ) / 100 = - 18 + 88 - 16 = 54 answer : b"
a ) 44 , b ) 54 , c ) 46 , d ) 47 , e ) 48
b
multiply(subtract(multiply(divide(subtract(const_100, 18), const_100), divide(add(const_100, 88), const_100)), const_1), const_100)
add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 38 now , the son age 5 years back was :
"explanation : let the son ' s present age be x years . then , ( 38 - x ) = x x = 19 . son ' s age 5 years back = ( 19 - 5 ) = 14 years . answer : a"
a ) 14 , b ) 19 , c ) 33 , d ) 38 , e ) 56
a
subtract(divide(38, const_2), 5)
divide(n0,const_2)|subtract(#0,n1)|
general
20 machines can do a work in 5 days . how many machines are needed to complete the work in 10 days ?
"required number of machines = 20 * 5 / 10 = 10 answer is a"
a ) 10 , b ) 15 , c ) 8 , d ) 12 , e ) 20
a
divide(multiply(20, 5), 10)
multiply(n0,n1)|divide(#0,n2)|
physics
in x - y plane , there is a right triangle abc ( ∠ b = 90 o ) . if the length of ac is 225 and the slope of line segment ac is 4 / 3 , what is the length of ab ?
"slope = change in vertical direction / change in horizontal direction = 4 / 3 change in vertical direction = 4 x = ab change in horizontal direction = 3 x = bc ab ^ 2 + bc ^ 2 = 225 ^ 2 16 x ^ 2 + 9 x ^ 2 = 225 ^ 2 25 x ^ 2 = 225 ^ 2 x ^ 2 = 25 * 9 ^ 2 x = 45 therefore ab = 45 * 4 = 180 answer : b"
a ) 12 , b ) 180 , c ) 24 , d ) 28 , e ) 40
b
multiply(4, divide(225, sqrt(add(power(4, const_2), power(3, const_2)))))
power(n2,const_2)|power(n3,const_2)|add(#0,#1)|sqrt(#2)|divide(n1,#3)|multiply(n2,#4)|
general
at company x , senior sales representatives visit the home office once every 18 days , and junior sales representatives visit the home office once every 12 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a sen...
"each 36 - day period , senior representatives visit the home office 2 times while junior representatives visit 3 times , thus 50 % more . the answer is d ."
a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 %
d
multiply(add(const_1, const_10), subtract(subtract(18, 12), const_1))
add(const_1,const_10)|subtract(n0,n1)|subtract(#1,const_1)|multiply(#0,#2)|
general
set a of 8 positive integers may have the same element and have 36 . and set b of 8 positive integers must have different elements and have 36 . when m and n are the greatest possible differences between 36 and other elements ’ sums in set a and set b , respectively , m - n = ?
this is maximum - minimum . hence , 36 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 29 and 36 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 8 . so , 29 - 8 = 21 . the correct answer is b .
a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26
b
subtract(36, add(add(8, const_2), 8))
add(n0,const_2)|add(n0,#0)|subtract(n1,#1)|
general
how many times digit 6 is used while writing numbers from 100 to 1100 ?
"there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' s...
a ) 648 , b ) 320 , c ) 252 , d ) 225 , e ) 26
b
add(add(divide(subtract(1100, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2))
add(const_1,const_10)|multiply(n0,const_2)|subtract(n2,n1)|divide(#2,const_10)|multiply(#0,#0)|add(#3,#4)|add(#5,#1)|
general
what is the total cost of 2 sandwiches at $ 3.49 each and 4 sodas at $ 0.87 each ?
"answer = d 2 * 3.49 + 4 * 0.87 = 2 ( 3.50 - 0.01 ) + 4 ( 1.00 - 0.13 ) = 7 + 4 - 0.02 - 0.52 = 11 - 0.54 = 10.46"
a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.46 , d ) $ 10.46 , e ) $ 11.85
d
add(multiply(2, 3.49), multiply(4, 0.87))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
general
gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 7 times as heavy as water ?
"g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 7 w 10 + 5 x = 7 ( x + 1 ) x = 3 / 2 ratio of gold with copper = 1 : 3 / 2 = 2 : 3 answer is e"
a ) 3 : 2 , b ) 1 : 2 , c ) 3 : 1 , d ) 5 : 2 , e ) 2 : 3
e
divide(subtract(7, 5), subtract(10, 7))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|
general
find the ratio in which rice at rs . 7.00 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg ?
"solution required ratio = 60 : 70 = 6 : 7 answer d"
a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 4 , d ) 6 : 7 , e ) none of these
d
divide(subtract(6.30, 5.70), subtract(7.00, 6.30))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|
other
the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 6 ) and ( 9 , 6 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ?
"if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 3 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 3 ) is 3 / 5 . the answer is d ."
a ) 3 , b ) 2 , c ) 1 / 4 , d ) 3 / 5 , e ) 5 / 6
d
divide(1, divide(add(subtract(9, 1), 6), 6))
subtract(n2,n0)|add(#0,n5)|divide(#1,n5)|divide(n0,#2)|
general
calculate how many seconds it will take for 4 bells to toll together again , given that they begin to toll together respectively at the intervals of 5 , 6 , 10 and 12 seconds . ?
"lcm of 5 , 6 , 10 and 12 is 60 lcm = 60 answer : b"
a ) 61 , b ) 60 , c ) 79 , d ) 70 , e ) 78
b
multiply(divide(10, const_2), multiply(5, 6))
divide(n3,const_2)|multiply(n1,n2)|multiply(#0,#1)|
gain
a water tank is one - fifth full . pipe a can fill a tank in 15 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how many minutes will it take to empty or fill the tank completely ?
"the combined rate of filling / emptying the tank = 1 / 15 - 1 / 6 = - 1 / 10 since the rate is negative , the tank will be emptied . a full tank would take 10 minutes to empty . since the tank is only one - fifth full , the time is ( 1 / 5 ) * 10 = 2 minutes the answer is a ."
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
a
multiply(divide(const_1, add(const_1, const_4)), divide(const_1, subtract(divide(const_1, 6), divide(const_1, 15))))
add(const_1,const_4)|divide(const_1,n1)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|divide(const_1,#4)|multiply(#3,#5)|
physics
a jar contains 4 black and 3 white balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ?
"p ( 1 st black , 2 nd white ) = 4 / 7 * 3 / 6 = 4 / 14 ; p ( 1 st white , 2 nd black ) = 3 / 7 * 4 / 6 = 4 / 14 . p = 4 / 14 + 4 / 14 = 4 / 7 . answer : c ."
a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 2
c
multiply(divide(4, add(4, 3)), divide(3, subtract(add(4, 3), const_1)))
add(n0,n1)|divide(n0,#0)|subtract(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|
probability
a can do a work in 3 days . b can do the same work in 6 days . if both a & b are working together in how many days they will finish the work ?
"a rate = 1 / 3 b rate = 1 / 6 ( a + b ) rate = ( 1 / 3 ) + ( 1 / 6 ) = 1 / 2 a & b finish the work in 2 days correct option is d"
a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6
d
divide(multiply(3, 6), add(3, 6))
add(n0,n1)|multiply(n0,n1)|divide(#1,#0)|
physics
in an it company , there are a total of 100 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ?
"you could pick 50 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 55 . c"
a ) 10 , b ) 50 , c ) 55 , d ) 35 , e ) 65
c
add(subtract(80, 100), subtract(50, 35))
subtract(n2,n0)|subtract(n1,n3)|add(#0,#1)|
general
in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 390 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ?
"the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 / 15 of the population . ( 4 / 15 ) * 390 = 104 the answer is c ."
a ) 84 , b ) 96 , c ) 104 , d ) 112 , e ) 120
c
multiply(divide(390, add(add(multiply(3, 2), 4), 5)), 4)
multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)|
other
a train 280 m long , running with a speed of 36 km / hr will pass a tree in
"sol . speed = ( 36 x 5 / 18 ) m / sec . = 10 m / sec . time taken = ( 280 x 1 / 10 ) sec = 28 sec answer d"
a ) 12 sec , b ) 15 sec , c ) 16 sec , d ) 28 sec , e ) none
d
multiply(divide(280, multiply(36, const_1000)), const_3600)
multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|
physics
for all real numbers v , an operation is defined by the equation v * = v - v / 3 . if ( v * ) * = 4 , then v =
"( v * ) * = ( v - v / 3 ) - ( v - v / 3 ) / 3 4 = 2 v / 3 - 2 v / 9 = 4 v / 9 v = 9 the answer is b ."
a ) 6 , b ) 9 , c ) 12 , d ) 15 , e ) 18
b
divide(divide(4, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3)))
divide(const_1,n0)|subtract(const_1,#0)|divide(n1,#1)|divide(#2,#1)|
general
a person buys an article at rs . 340 . at what price should he sell the article so as to make a profit of 10 % ?
"cost price = rs . 340 profit = 10 % of 340 = rs . 34 selling price = cost price + profit = 340 + 34 = 374 answer : c"
a ) 600 , b ) 882 , c ) 374 , d ) 356 , e ) 521
c
add(340, multiply(340, divide(10, const_100)))
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
on dividing 12401 by a certain number , we get 76 as quotient and 13 as remainder . what is the divisor ?
"divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 12401 - 13 ) / 76 = 163 answer ( a )"
a ) 163 , b ) 785 , c ) 852 , d ) 456 , e ) 852
a
divide(subtract(12401, 13), 76)
subtract(n0,n2)|divide(#0,n1)|
general
the average of the marks of 10 students in a class is 40 . if the marks of each student are doubled , find the new average ?
"sum of the marks for the 10 students = 10 * 40 = 400 . the marks of each student are doubled , the sum also will be doubled . the new sum = 400 * 2 = 800 . so , the new average = 800 / 10 = 80 . answer : c"
a ) 78 , b ) 56 , c ) 80 , d ) 27 , e ) 40
c
multiply(40, const_2)
multiply(n1,const_2)|
general
how much interest will $ 10,000 earn in 9 months at an annual rate of 6 % ?
"soln : - 9 months = 3 / 4 of year ; 6 % = 6 / 100 = 3 / 50 ; $ 10,000 ( principal ) * 3 / 50 ( interest rate ) * 3 / 4 ( time ) = $ 450 . answer : c"
a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 550 , e ) $ 650
c
multiply(multiply(power(const_100, const_2), divide(const_4.0, const_100)), divide(const_3, 6))
divide(const_3,const_4)|divide(n2,const_100)|power(const_100,const_2)|multiply(#1,#2)|multiply(#0,#3)|
gain
what is the sum of the greatest common factor and the lowest common multiple of 72 and 36 ?
"prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 72 = 2 x 2 x 2 x 3 x 3 gcf = 36 lcm = 72 sum = 108 ans : d"
a ) 120 , b ) 150 , c ) 145 , d ) 108 , e ) 112
d
divide(multiply(72, 36), const_4)
multiply(n0,n1)|divide(#0,const_4)|
general
at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 19.6 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ?
"let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 19.6 x ^ 2 = 1.96 x = 1.4 at 4 : 00 pm , the amount of bacteria was 10 ( 1.4 ) = 14 grams . the answer is b ."
a ) 13.7 , b ) 14.0 , c ) 14.3 , d ) 14.6 , e ) 14.9
b
multiply(sqrt(divide(19.6, 10.0)), 10.0)
divide(n5,n2)|sqrt(#0)|multiply(#1,n2)|
general
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 350 . find the total worth of the stock ?
"let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 350 = > x = 17500 answer : d"
a ) 20029 , b ) 20000 , c ) 20289 , d ) 17500 , e ) 20026
d
divide(350, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100))))
divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)|
gain
farm tax is levied on the 60 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :
"this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 60 ) = 20.833 % and...
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) none
e
divide(multiply(multiply(divide(480, 3840), const_100), const_100), 60)
divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|
general
if 3 spiders make 3 webs in 3 days , then 1 spider will make 1 web in how many days ?
"let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 3 webs 3 : 1 1 x 3 x x = 3 x 1 x 3 = > x = 3 answer is e"
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3
e
multiply(1, 3)
multiply(n0,n3)|
physics
three machines , each working at the same constant rate , together can complete a certain job in 18 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 9 days ?
"rate of one machine = 1 job / ( 18 * 3 ) days let x = number of machines needed to complete the job in 9 days 1 / ( 3 * 18 ) * 9 * x = 1 job x = 6 6 - 3 = 3 answer : a"
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 8
a
subtract(divide(multiply(18, add(const_4, const_1)), 9), add(const_4, const_1))
add(const_1,const_4)|multiply(n0,#0)|divide(#1,n1)|subtract(#2,#0)|
physics
light glows for every 30 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am .
"time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 30 ) = 165 times . answer : e"
a ) 380 times , b ) 142 times , c ) 382 times , d ) 392 times , e ) 165 times
e
divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 30)
add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)|
general
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 2 : 5 . the first number is :
"( 3 x - 9 ) : ( 5 x - 9 ) = 2 : 5 x = 11 = > 3 x = 33 answer : a"
a ) a ) 33 , b ) b ) 98 , c ) c ) 34 , d ) d ) 35 , e ) e ) 62
a
add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5))))
multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
other
a man is 35 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is :
"let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 35 ) + 2 = 2 ( x + 2 ) x + 37 = 2 x + 4 x = 33 . answer : e"
a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 33 years
e
divide(subtract(35, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))
multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|
general
p and q invested in a business . the profit earned was divided in the ratio 4 : 6 . if p invested rs 60000 , the amount invested by q is
"let the amount invested by q = q 60000 : q = 4 : 6 ⇒ 60000 × 6 = 4 q ⇒ q = ( 60000 × 6 ) / 4 = 90000 answer is c ."
a ) 30000 , b ) 50000 , c ) 90000 , d ) 20000 , e ) 60000
c
multiply(divide(60000, 4), 6)
divide(n2,n0)|multiply(n1,#0)|
gain
on charlie ' s cell phone plan , he signed up for 8 gb of data per 4 week period . if he went over the 8 gb per month , it would cost him an extra $ 10.00 per gb . in the first week , he used 2 gb ; in the 2 nd week , he used 3 gb ; in the 3 rd week , he used 5 gb ; and in the 4 th week , he used a whopping 10 gb . how...
to determine how much extra he would need to pay , you have to add all 4 weeks of gb usage ( 2 + 3 + 5 + 10 = 20 ) . to determine the extra amount of gb usage , you have to subtract the the original amount of gb ( 8 ) allowed on his plan . 20 - 8 = 12 to determine the extra cost on his cell phone plan , you have to mul...
a ) $ 120 , b ) $ 40 , c ) $ 80 , d ) $ 100 , e ) $ 140
a
multiply(subtract(add(add(add(2, 3), 5), 10), 8), 10)
add(n4,n6)|add(n8,#0)|add(n3,#1)|subtract(#2,n0)|multiply(n3,#3)
general
a pipe can fill a cistern in 44 minutes . find the time in which 1 / 11 part of the cistern will be filled ?
"full cistern filled in = 44 minutes 1 / 11 part filled in = 44 * 1 / 11 = 4 minutes answer is a"
a ) 4 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min
a
multiply(44, divide(1, 11))
divide(n1,n2)|multiply(n0,#0)|
physics
0.0003 x 0.3 = ?
"3 x 3 = 9 . sum of decimal places = 5 0.0003 x 0.3 = 0.00009 answer : option b"
a ) 0.0009 , b ) 9.0 e - 05 , c ) 0.09 , d ) 0.009 , e ) 9.0
b
multiply(divide(0.0003, 0.3), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 π · r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams e...
"weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )"
a ) e = 16 , b ) e = 32 , c ) e = 64 , d ) 128 , e ) 512
b
multiply(8, 4)
multiply(n0,n3)|
geometry
a train running at the speed of 54 km / hr crosses a pole in 7 seconds . find the length of the train .
"explanation : speed = 54 * ( 5 / 18 ) m / sec = 15 m / sec length of train ( distance ) = speed * time = 15 * 7 = 105 meter option b"
a ) 150 meter , b ) 105 meter , c ) 140 meter , d ) 135 meter , e ) none of these
b
multiply(divide(multiply(54, const_1000), const_3600), 7)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
what will be the fraction of 25 %
"explanation : it will 25 * 1 / 100 = 1 / 4 option a"
a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 1 / 11 , e ) none of above
a
divide(circle_area(divide(25, const_2)), const_2)
divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)|
gain
what is difference between biggest and smallest fraction among 2 / 3 , 3 / 4 , 4 / 5 and 5 / 4
"explanation : 2 / 3 = . 66 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 4 = 1.25 so biggest is 5 / 4 and smallest is 2 / 3 their difference is 5 / 4 - 2 / 3 = 7 / 12 option c"
a ) 2 / 5 , b ) 3 / 5 , c ) 7 / 12 , d ) 1 / 7 , e ) none of these
c
subtract(divide(4, 5), divide(2, 3))
divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)|
general
an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 4 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ?
let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 4 y and y . 2 ( 3 x + 2 x ) = 2 ( 4 y + y ) 5 x = 5 y x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 4 y * y = 6 x ^ 2 : 4 y ^ 2 = 6 x ^ 2 : 4 x ^ 2 = 6 : 4 = 3 : 2 the answer is b .
['a ) 5 : 8', 'b ) 3 : 2', 'c ) 6 : 1', 'd ) 8 : 7', 'e ) 5 : 6']
b
divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), 4)))
add(n2,n3)|rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|divide(#2,const_2)|divide(#3,#0)|multiply(n2,#4)|rectangle_area(#4,#5)|divide(#1,#6)
geometry
if the sides of a triangle are 26 cm , 24 cm and 20 cm , what is its area ?
"the triangle with sides 26 cm , 24 cm and 20 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 24 * 20 = 240 cm 2 answer : e"
a ) 120 , b ) 772 , c ) 288 , d ) 266 , e ) 240
e
divide(multiply(24, 20), const_2)
multiply(n1,n2)|divide(#0,const_2)|
geometry
a metallic sheet is of rectangular shape with dimensions 48 m x 38 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is :
"l = ( 48 - 16 ) m = 32 m , b = ( 38 - 16 ) m = 22 m , h = 8 m . volume of the box = ( 32 x 22 x 8 ) m 3 = 5632 m 3 . answer : option b"
a ) 4830 , b ) 5632 , c ) 6420 , d ) 7500 , e ) 8960
b
volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(38, multiply(8, const_2)), 8)
multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|
geometry
it takes 40 identical printing presses 9 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ?
"40 printing presses can do 1 / 9 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 9 = 1 / 12 of the job each hour . the answer is d ."
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
d
divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 9), 30))
divide(const_1000,const_2)|multiply(#0,const_1000)|divide(#1,n0)|divide(#2,n1)|multiply(n3,#3)|divide(#1,#4)|
physics
the average age of 15 students of a class is 16 years . out of these , the average age of 6 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ?
"age of the 15 th student = [ 15 * 16 - ( 14 * 6 + 16 * 9 ) ] = ( 240 - 228 ) = 12 years . answer : c"
a ) 10 years , b ) 11 years , c ) 12 years , d ) 13 years , e ) 14 years
c
subtract(multiply(15, 15), add(multiply(6, 14), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 6 inches wide on each side , the total area would have been ( m + 200 ) square inches . what is the perimeter of the photograph , in inches ?
"let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 12 ) ( y + 12 ) = m and so ( 2 ) xy + 12 x + 12 y + 144 = m + 200 let ' s subtract equation ( 1 ) from equation ( 2 ) . 10 x + 10 y + 140 = 200 2 x + 2 y = 12 , which is the perimeter of the photog...
a ) 8 , b ) 12 , c ) 16 , d ) 20 , e ) 24
b
divide(subtract(200, subtract(power(multiply(6, const_2), const_2), power(multiply(1, const_2), const_2))), const_2)
multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)|
geometry
how many numbers between 100 and 600 are divisible by 2 , 3 , and 7 together ?
"explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 600 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 600 : 600 / 42 = 14.28 hence , 42 * 14 = 588 is the last number divisible by 42 within 60...
a ) 11 , b ) 12 , c ) 14 , d ) 16 , e ) 18
b
subtract(divide(600, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7)))
multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)|
general
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 5 hours to fill the tank . the leak can drain all the water of the tank in ?
work done by the tank in 1 hour = ( 1 / 2 - 2 1 / 5 ) = 1 / 22 leak will empty the tank in 22 hrs . answer : c
a ) 17 hr , b ) 19 hr , c ) 22 hr , d ) 24 hr , e ) 26 hr
c
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 5), 1), 5))))
divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)
physics
the salary of a labourer increase consistently by 40 % per year . if his present salary is rs . 3000 then what will his salary be after 3 years ?
explanation : salary = 3000 x 1.4 x 1.4 x 1.4 = 8232 answer : option a
a ) 8232 , b ) 3282 , c ) 4282 , d ) 8242 , e ) 2382
a
multiply(3000, power(add(const_1, divide(40, const_100)), 3))
divide(n0,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n1,#2)
gain
in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 30 % of the people who do not have the job of their choice have a university diploma . if 40 % of the people have the job of their choice , what percent of the people have a university diploma ?
setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = 10 % job of choice total = 40 % not job of choice with diploma = . 3 x not job of choice w / o diploma = . 7 x total not job of choice = x total with diploma total without diploma total citize...
a ) 35 % , b ) 48 % , c ) 55 % , d ) 65 % , e ) 75 %
b
add(divide(multiply(30, subtract(const_100, 40)), const_100), subtract(40, 10))
subtract(const_100,n2)|subtract(n2,n0)|multiply(n1,#0)|divide(#2,const_100)|add(#3,#1)
gain
in a house a hall is 20 m long , 15 m wide and 5 m high . its interior has to be covered with mat . what will be the total expenditure if it costs rs . 50 per square m ?
"length ( l ) = 20 m , breadth ( b ) = 15 m and height ( h ) = 5 m total area of the hall to be covered with mat = 2 ( lb + bh + hl ) = 2 ( 20 * 15 + 15 * 5 + 5 * 20 ) = 2 ( 300 + 75 + 100 ) = 2 * 475 = 950 sq . m total expenditure = 50 * 950 = rs . 47500 e"
a ) s . 43000 , b ) s . 50000 , c ) s . 57000 , d ) s . 60000 , e ) s . 47500
e
add(add(multiply(multiply(multiply(20, 15), const_2), 50), multiply(multiply(multiply(20, 5), const_2), 50)), multiply(multiply(multiply(15, 5), const_2), 50))
multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(n3,#3)|multiply(n3,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#8)|
general
there is a 55 cm line marked at each centimeter and an insect is placed at every centimeter . 9 frogs are trained to jump a constant distance . the first one jumps 2 cm in every leap , the second one jumps 3 cm and so on until the 9 th one jumps 10 cm in every leap and they eat any insect that is available at that spot...
only the prime numbers greater than 10 and less than 55 were left . that is 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , and 53 . the total is 12 . the answer is e .
a ) 0 , b ) 3 , c ) 6 , d ) 9 , e ) 12
e
add(10, 2)
add(n2,n5)
physics
jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 58,000 after completion of one year . what will be madhu ' s share of profit ?
30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 58,000 i . e . rs . 29,000 answer : d
a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 30,000 , d ) rs . 29,000 , e ) none of these
d
multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(const_2, const_3), const_2)), const_100), multiply(multiply(add(const_2, const_3), const_100), const_100)), divide(divide(multiply(add(const_2, const_3), const_2), const_2), multiply(const_4, const_3)))
add(const_2,const_3)|multiply(n1,const_4)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#0,const_100)|divide(#3,const_2)|multiply(#1,#3)|multiply(#4,const_100)|divide(#5,#2)|multiply(#6,const_100)|add(#9,#7)|multiply(#10,#8)
gain
there is enough provisions for 600 men in an army camp for 35 days . if there were 300 men less , how long will the provision last ?
"exp : we have , m 1 d 1 = m 2 d 2 600 * 35 = 300 * d 2 d 2 = 600 * 35 / 300 = 70 days . answer : d"
a ) 40 days , b ) 50 days , c ) 60 days , d ) 70 days , e ) 80 days
d
divide(multiply(600, 35), 300)
multiply(n0,n1)|divide(#0,n2)|
physics
if 8 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ?
"parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 8 c 2 for vertical 8 c 2 total parallelogram is 10 c 2 * 8 c 2 = 28 * 28 = 784 answer : a"
a ) 784 , b ) 763 , c ) 120 , d ) 160 , e ) 1260
a
multiply(divide(multiply(8, subtract(8, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2))
subtract(n0,const_1)|subtract(n1,const_1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,const_2)|divide(#3,const_2)|multiply(#4,#5)|
physics
there are 100 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off ...
"number of switches = 100 number of switches turned on by a : 3 , 6 , . . . 99 = 33 number of switches turned on by b : 5 , 10 , . . . . 100 = 20 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b as they are turned...
a ) 13 , b ) 41 , c ) 47 , d ) 48 , e ) 53
b
subtract(add(floor(divide(100, const_3)), floor(divide(100, add(const_1, const_4)))), multiply(floor(divide(100, multiply(const_3, add(const_1, const_4)))), const_2))
add(const_1,const_4)|divide(n0,const_3)|divide(n0,#0)|floor(#1)|multiply(#0,const_3)|divide(n0,#4)|floor(#2)|add(#3,#6)|floor(#5)|multiply(#8,const_2)|subtract(#7,#9)|
other
at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 14 charts have been brought to the meeting , how man...
say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 14 add both : 3 a + 3 b = 24 so a + b = 8 total number of people = 8 c
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
c
divide(add(14, 10), add(2, 1))
add(n4,n5)|add(n0,n1)|divide(#0,#1)
general
30 % people of a village in sri lanka died by bombardment , 25 % of the remainder left the village on account of fear . if now the population is reduced to 6695 , how much was it in the beginning ?
"x * ( 70 / 100 ) * ( 75 / 100 ) = 3515 x = 6695 answer : c"
a ) 7000 , b ) 6700 , c ) 6695 , d ) 7645 , e ) 6575
c
floor(divide(6695, multiply(divide(subtract(const_100, 30), const_100), divide(subtract(const_100, 25), const_100))))
subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)|
general
i gain 70 paise on rs . 70 . my gain percent is
"solution gain % = ( 0.70 / 70 × 100 ) % = 1 % . answer b"
a ) 0.1 % , b ) 1 % , c ) 7 % , d ) 10 % , e ) none
b
divide(multiply(divide(70, const_100), const_100), 70)
divide(n0,const_100)|multiply(#0,const_100)|divide(#1,n0)|
gain
50 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 4 , then the rise in the water level in the tank will be :
explanation : total volume of water displaced = ( 4 x 50 ) = 200 rise in water level = = 0.25 m = 25 cm answer : b ) 25 cm
a ) 22 , b ) 25 , c ) 32 , d ) 99 , e ) 88
b
multiply(divide(multiply(50, 4), multiply(40, 20)), const_100)
multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)|multiply(#2,const_100)
physics