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two pipes can fill a tank in 10 minutes and 15 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ?
"part of the filled by all the three pipes in one minute = 1 / 10 + 1 / 15 - 1 / 45 = ( 9 + 6 - 2 ) / 90 = 13 / 90 so , the tank becomes full in 8 minutes . answer : b"
a ) 30 minutes , b ) 8 minutes , c ) 15 minutes , d ) 10 minutes , e ) 12 minutes
b
subtract(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 45))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|
physics
dean winchester has got a long wooden stock of size 60 feet , he need to cut small stock of size 1 feet long using his axe . he takes 5 minutes to cut one small stock ( 1 feet ) , how long will he take to make 60 such small stocks ?
solution : 295 minutes . when he cut 59 stocks , the 60 th stock will remain . 59 * 5 = 295 minutes . answer c
a ) 296 minutes . , b ) 297 minutes . , c ) 295 minutes . , d ) 294 minutes . , e ) none
c
multiply(60, 5)
multiply(n0,n2)
physics
the difference between the squares of two consecutive odd integers is a square < 1000 whose root equals the combined sum of the digits of the two integers . what is the sum of the digits x of the larger integer ?
really hard . but let ' s start with few concepts : 1 . square of an odd integer will be odd . 2 . difference of odd and odd will be even . 3 . any odd integer can be expressed as difference of two squares . 4 . an even integer can be expressed as difference of two squares only if that even integer is a multiple of 4 ....
a ) x = 2 , b ) x = 5 , c ) x = 8 , d ) 10 , e ) 11
a
add(divide(multiply(divide(power(subtract(floor(add(sqrt(1000), const_1)), const_12), const_2), power(const_2, const_3)), const_2), const_100), const_1)
power(const_2,const_3)|sqrt(n0)|add(#1,const_1)|floor(#2)|subtract(#3,const_12)|power(#4,const_2)|divide(#5,#0)|multiply(#6,const_2)|divide(#7,const_100)|add(#8,const_1)
general
if c is 25 % of a and 50 % of b , what percent of a is b ?
"answer = b 25 a / 100 = 50 b / 100 b = 25 a / 50 = 50 a / 100 = 50 %"
a ) 2.5 % , b ) 50 % , c ) 25 % , d ) 35 % , e ) 250 %
b
multiply(divide(divide(25, const_100), divide(50, const_100)), const_100)
divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|
gain
a metallic cuboid of dimensions 27 x 18 x 12 m is melted . find the minimum amount of the volume of molten metal which should be added to mould it into a cube whose edge is ' a ' cm where a is an integer .
volume of cuboid = 27 * 18 * 12 = ( 27 * 27 * 8 ) = ( 3 * 3 * 2 ) ^ 3 = 18 ^ 3 so no amount of the volume of molten metal should be added to mould it into a cube whose edge is ' a ' cm where a is 18 m . answer : c
['a ) 16 m', 'b ) 17 m', 'c ) 18 m', 'd ) 19 m', 'e ) 15 m']
c
cube_edge_by_volume(multiply(multiply(27, 18), 12))
multiply(n0,n1)|multiply(n2,#0)|cube_edge_by_volume(#1)
physics
a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 61 , how many questions did student a answer correctly ?
"let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 61 ( subtracting twice of incorrect from correct ) 3 x = 261 x = 87 answer : e"
a ) 55 , b ) 60 , c ) 73 , d ) 81 , e ) 87
e
subtract(100, divide(subtract(100, 61), const_3))
subtract(n0,n1)|divide(#0,const_3)|subtract(n0,#1)|
general
if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 600 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 615 . what is the value of x ?
"simple way to solve this question is to use options . from si , we know that x * y = 30,000 . now , put the value of x = 6000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 6000 + 5 % of 6000 = 6300 . 2 nd year , amount = 6300 + 5 % of 6300 = 6615 we can see after 2 years interest = 6615 - 6...
a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000
b
divide(power(divide(600, const_2), const_2), subtract(615, 600))
divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)|
gain
a store purchased a pair of pants for $ 210 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a store owner decided to decrease the selling price by 20 percent . what was the store ' s gross profit on this sale ?
sale price ( sp ) = 210 + markup ( mp ) - - > mp = sp - 210 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 210 3 sp / 4 = 210 sp = 280 now a discount of 20 % is given so new sp is . 8 * 280 = 224 profit = 224 - 210 = 14 answer is a
a ) $ 14 , b ) $ 2 , c ) $ 10 , d ) $ 6 , e ) $ 8
a
divide(divide(multiply(210, 20), const_100), const_3)
multiply(n0,n2)|divide(#0,const_100)|divide(#1,const_3)
general
the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 400 per sq . metre .
"solution area of the floor = ( 5.5 Γ— 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 400 Γ— 20.625 ) = 8250 . answer b"
a ) rs . 15000 , b ) rs . 8250 , c ) rs . 15600 , d ) rs . 16500 , e ) none of these
b
multiply(400, multiply(5.5, 3.75))
multiply(n0,n1)|multiply(n2,#0)|
physics
find 125 / 14 * 6
"answer = 125 / 14 * 6 = 125 / 84 = 1.488 option c is correct"
a ) 2.4658 , b ) 6.398 , c ) 1.488 , d ) 5.687 , e ) 9.652
c
divide(125, 14)
divide(n0,n1)|
general
if a 2 - b 2 = 5 and a * b = 2 , find a 4 + b 4 .
"a 2 - b 2 = 5 : given a 4 + b 4 - 2 a 2 b 2 = 52 : square both sides and expand . a * b = 2 : given a 2 b 2 = 22 : square both sides . a 4 + b 4 - 2 ( 4 ) = 25 : substitute a 4 + b 4 = 33 correct answer e"
a ) 50 , b ) 72 , c ) 25 , d ) 92 , e ) 33
e
add(power(5, 2), multiply(power(2, 2), 2))
power(n3,n0)|power(n2,n0)|multiply(#0,n0)|add(#2,#1)|
general
if 1 + 2 + 3 + . . . + n = n ( n + 1 ) , then 3 ( 1 + 3 + 5 + . . . . + 79 ) = ?
"explanation : to solve this use the formula of ap , sn = ( n / 2 ) ( a + l ) . . . . . . . . . . . . . . . . ( 1 ) to find n , use = > tn = a + ( n - 1 ) d = > 79 = 1 + ( n - 1 ) 2 = > n = 40 use value of n in ( 1 ) then , sn = ( 40 / 2 ) ( 1 + 79 ) = 1600 ans : - 3 ( sn ) = 4800 answer : d"
a ) 4500 , b ) 4250 , c ) 4350 , d ) 4800 , e ) 4150
d
multiply(subtract(divide(multiply(79, add(79, 1)), 2), multiply(divide(subtract(79, 1), 2), add(divide(subtract(79, 1), 2), 1))), 3)
add(n8,n0)|subtract(n8,n0)|divide(#1,n1)|multiply(n8,#0)|add(#2,n0)|divide(#3,n1)|multiply(#4,#2)|subtract(#5,#6)|multiply(#7,n2)|
general
a river boat leaves silver town and travels upstream to gold town at an average speed of 4 kilometers per hour . it returns by the same route at an average speed of 7 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ?
"pick a number which is lcm of 7 and 4 = 18 . upstream time = 28 / 4 = 7 hrs downstream time = 28 / 7 = 4 hrs total time = 11 hrs total distance = 56 average speed = 56 / 11 = 5.1 km / hr"
a ) 5.1 , b ) 7.1 , c ) 7.2 , d ) 7.5 , e ) 8.0
a
divide(multiply(multiply(4, 7), const_2), add(4, 7))
add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|divide(#2,#0)|
physics
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 25 miles per gallon , and from town b to town c , the car averaged 30 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through ...
step 1 ) took lcm of 25 and 30 . . came as 150 . step 2 ) 150 distance between b to c . . . do 150 / 30 hence 5 gallons used step 3 ) twice distance . . hence 150 * 2 = 300 . . . do as above . . 300 / 25 = 12 gallons used step 4 ) total gallons . . 5 + 12 = 17 gallons step ) total miles = 150 + 300 = 450 miles hence . ...
a ) 24 , b ) 25 , c ) 23 , d ) 22 , e ) 26.5
e
divide(add(multiply(30, const_10), divide(multiply(30, const_10), const_2)), add(divide(multiply(30, const_10), 25), divide(divide(multiply(30, const_10), const_2), 30)))
multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)
general
if w , x , y and z are distinct odd positive integers , then the maximum value of the expression ( w ^ 2 + x ^ 2 ) * ( y ^ 2 + z ^ 2 ) can be divisible by ?
any odd number square will give odd number only . . similarly wen we add 2 odd numbers we will get even number . . wen we multiply 2 even owe will get even nuber . . so the result of the above expression is an even number only . . hence it is divisible by 2 . . . answer : a
a ) 2 , b ) 3 , c ) 5 , d ) 8 , e ) 10
a
divide(multiply(add(2, 2), 2), multiply(2, 2))
add(n0,n0)|multiply(n0,n0)|multiply(n0,#0)|divide(#2,#1)
general
john found that the average of 15 numbers is 40 . if 10 is added to each number then the mean of number is ?
"( x + x 1 + . . . x 14 ) / 15 = 40 50 option a"
a ) 50 , b ) 60 , c ) 75 , d ) 80 , e ) 85
a
add(40, 10)
add(n1,n2)|
general
if a ' s salary is 20 % less than b ' s salary , by how much percent is b ' s salary more than a ' s ?
solution required percentage β€Ή = β€Ί [ 20 / ( 100 - 20 ) x 100 ] % β€Ή = β€Ί 25 % answer d
a ) 5 % , b ) 10 % , c ) 15 % , d ) 25 % , e ) none
d
divide(subtract(const_100, subtract(const_100, 20)), divide(subtract(const_100, 20), const_100))
subtract(const_100,n0)|divide(#0,const_100)|subtract(const_100,#0)|divide(#2,#1)
general
what will be the difference between simple interest and compound interest @ 10 % per annum on a sum of rs . 1000 after 3 years ?
"explanation : s . i . = rs . [ ( 1000 x 10 x 3 ) / 100 ] = rs . 300 c . i . = rs . [ 1000 x ( 1 + ( 10 / 100 ) ) 3 - 1000 ] = rs . 331 difference = rs . ( 331 - 300 ) = rs . 31 answer : e"
a ) 22 , b ) 77 , c ) 26 , d ) 19 , e ) 31
e
subtract(subtract(multiply(power(add(divide(10, const_100), const_1), 3), 1000), 1000), divide(multiply(3, multiply(10, 1000)), const_100))
divide(n0,const_100)|multiply(n0,n1)|add(#0,const_1)|multiply(n2,#1)|divide(#3,const_100)|power(#2,n2)|multiply(n1,#5)|subtract(#6,n1)|subtract(#7,#4)|
gain
the mean of 25 values was 190 . it was detected on rechecking that one value 165 was wrongly copied as 130 for the computation of the mean . find the correct mean .
"corrected mean = 190 Γ— 25 βˆ’ 130 + 165 / 25 = 4750 + 35 / 25 = 4785 / 25 = 191.4 answer d"
a ) 151 , b ) 149 , c ) 152 , d ) 191.4 , e ) none of the above
d
divide(add(multiply(25, 190), subtract(165, 130)), 25)
multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)|
general
a man can swim in still water at 5 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?
m = 5 s = x ds = 5 + x us = 5 - x 5 + x = ( 5 - x ) 2 5 + x = 10 - 2 x 3 x = 5 x = 1.667 answer : b
a ) 1.78 , b ) 1.667 , c ) 1.15 , d ) 1.5 , e ) 1.2
b
divide(5, const_3)
divide(n0,const_3)
general
in an office , 10 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ?
"( 10 / 100 ) workers = 16 = > number of workers = 160 ( 10 / 100 ) * workers = x + 16 = > x = 16 answer a"
a ) 16 , b ) 64 , c ) 50 , d ) 144 , e ) 160
a
divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4))
divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)|
gain
micheal and adam can do together a piece of work in 20 days . after they have worked together for 11 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately .
rate of both = 1 / 20 together they do = 1 / 20 * 11 = 11 / 20 left work = 1 - 11 / 20 = 9 / 20 adam completes 9 / 20 work in 10 day so he took 10 * 20 / 9 = 200 / 9 days to complete the left work alone . thus the rate of adam is 9 / 200 rate of micheal = 1 / 20 - 9 / 200 = 1 / 200 thus micheal takes 200 days to comple...
a ) 80 days , b ) 100 days , c ) 120 days , d ) 110 days , e ) 200 days
e
inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 11))), 10))))
inverse(n0)|multiply(n1,#0)|subtract(const_1,#1)|inverse(#2)|multiply(n2,#3)|inverse(#4)|subtract(#0,#5)|inverse(#6)
physics
a train crosses a platform of 250 m in 20 sec , same train crosses another platform of length 640 m in 40 sec . then find the length of the train ?
"length of the train be β€˜ x ’ x + 250 / 20 = x + 640 / 40 2 x + 500 = x + 640 x = 140 m answer : d"
a ) 150 m , b ) 180 m , c ) 152 m , d ) 140 m , e ) 160 m
d
subtract(multiply(640, divide(20, divide(20, const_3))), multiply(250, divide(40, divide(20, const_3))))
divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)|
physics
the difference between the ages of two persons is 11 years . ten years ago , the elder one was twice as old as the younger one . the present age of the elder person is
"let their ages of x years and ( x + 11 ) years respectively . then , ( x + 11 ) - 10 = 2 ( x - 10 ) ⇔ x + 1 = 2 x - 20 ⇔ 3 x = 21 . ∴ present age of the elder person = ( 7 + 11 ) = 18 years . answer : 18 years"
a ) 18 , b ) 19 , c ) 20 , d ) 17 , e ) 16
a
add(subtract(11, subtract(11, add(const_3, const_2))), multiply(subtract(11, add(const_3, const_2)), const_2))
add(const_2,const_3)|subtract(n0,#0)|multiply(#1,const_2)|subtract(n0,#1)|add(#2,#3)|
general
a walks at 10 kmph and 3 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ?
"suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 3 hours . x / 10 - x / 20 = 3 x = 60 km answer is c"
a ) 100 km , b ) 150 km , c ) 60 km , d ) 120 km , e ) 200 km
c
multiply(3, 20)
multiply(n1,n2)|
physics
what would be the length of the diagonal of a square plot whose area is equal to the area of a rectangular plot of 45 m length and 40 m width ?
a 2 = 45 Γ— 40 = 1800 ∴ a = √ 1800 = 30 √ 2 ∴ diagonal of the square = √ 2 a = √ 2 Γ— 30 √ 2 = 30 Γ— 2 = 60 m answer b
['a ) 42.5 m', 'b ) 60 m', 'c ) 4800 m', 'd ) data inadequate', 'e ) none of these']
b
diagonal(square_edge_by_area(rectangle_area(45, 40)), square_edge_by_area(rectangle_area(45, 40)))
rectangle_area(n0,n1)|square_edge_by_area(#0)|diagonal(#1,#1)
geometry
a goods train runs at the speed of 72 km / hr and crosses a 80 m long platform in 26 sec . what is the length of the goods train ?
"speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 80 ) / 26 = 20 x = 440 m . answer : b"
a ) 382 , b ) 440 , c ) 278 , d ) 270 , e ) 881
b
subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 80)
divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)|
physics
what is the total cost of 2 sandwiches at $ 1.49 each and 4 sodas at $ 0.87 each ?
answer = b 2 * 1.49 + 4 * 0.87 = 2 ( 1.50 - 0.01 ) + 4 ( 1.00 - 0.13 ) = 3 + 4 - 0.02 - 0.52 = 7 - 0.54 = 6.46
a ) $ 3.36 , b ) $ 6.46 , c ) $ 8.46 , d ) $ 10.08 , e ) $ 11.85
b
add(multiply(2, 1.49), multiply(4, 0.87))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)
general
population of a city in 20004 was 1500000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007
"required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 3251625 / 2 e"
a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 3251625 / 2
e
multiply(1500000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100))))
divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)|
gain
how many prime numbers between 1 and 100 are factors of 7,150 ?
"make prime factorization of 7,150 - - > 7,150 = 2 βˆ— 5 ^ 2 βˆ— 11 βˆ— 13 so , 4 prime numbers between 1 and 100 ( namely 2 , 5 , 11 , and 13 ) are factors of 7,150 . answer : d ."
a ) one , b ) two , c ) three , d ) four , e ) five
d
multiply(const_4, 1)
multiply(n0,const_4)|
other
a firm is comprised of partners and associates in a ratio of 2 : 61 . if 60 more associates were hired , the ratio of partners to associates would be 4 : 137 . how many partners are currently in the firm ?
"the ratio 2 : 61 = 4 : 122 so the ratio changed from 4 : 122 to 4 : 137 . 137 - 122 = 15 which is 1 / 4 of the increase in 60 associates . the ratio changed from 16 : 488 to 16 : 548 . thus the number of partners is 16 . the answer is c ."
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20
c
multiply(divide(60, subtract(multiply(137, 2), 61)), 2)
multiply(n0,n4)|subtract(#0,n1)|divide(n2,#1)|multiply(n0,#2)|
other
5 , 7 , 11 , 18 , 33 , ( . . . )
"explanation : 5 5 Γ— 2 - 3 = 7 7 Γ— 2 - 3 = 11 11 Γ— 2 - 3 = 18 18 Γ— 2 - 3 = 33 33 Γ— 2 - 3 = 63 answer : option d"
a ) 22 , b ) 35 , c ) 27 , d ) 63 , e ) 25
d
subtract(negate(18), multiply(subtract(7, 11), divide(subtract(7, 11), subtract(5, 7))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
if the radius of a circle that centers at the origin is 5 , how many w points on the circle have integer coordinates ?
"i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : w = c"
a ) 4 , b ) 8 , c ) 12 , d ) 15 , e ) 20
c
subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2))
add(n0,const_4)|multiply(n0,n0)|power(const_2,const_2)|subtract(#1,#0)|subtract(#3,#2)|
geometry
how many multiples of 7 are there between 14 and 252 ?
"it should be mentioned whether 14 and 252 are inclusive . if 14 and 252 are inclusive , then the answer is ( 252 - 14 ) / 7 + 1 = 35 . if 14 and 252 are not inclusive , then the answer is ( 245 - 21 ) / 7 + 1 = 33 . since oa is a , then we have not inclusive case ."
a ) 33 , b ) 46 , c ) 59 , d ) 66 , e ) 80
a
add(divide(subtract(252, 14), 7), const_1)
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
general
solve x 2 Γ’ € β€œ 3 x Γ’ € β€œ 4 = 0 .
this one factors easily : x 2 Γ’ € β€œ 3 x Γ’ € β€œ 4 = 0 ( x + 1 ) ( x Γ’ € β€œ 4 ) = 0 x + 1 = 0 or x Γ’ € β€œ 4 = 0 x = Γ’ € β€œ 1 or x = 4 the solution is x = Γ’ € β€œ 1 , 4 answer b
a ) 14 , b ) - 14 , c ) 23 , d ) 2 , - 3 , e ) 34
b
multiply(4, 2)
multiply(n0,n2)
general
the height of two right circular cones are in the ratio 1 : 2 and their perimeters of their bases are in the ratio 3 : 4 , the ratio of their volume is ?
9 : 32 answer : c
['a ) 9 : 38', 'b ) 9 : 31', 'c ) 9 : 32', 'd ) 9 : 39', 'e ) 9 : 35']
c
multiply(divide(1, 2), power(divide(3, 4), const_2))
divide(n0,n1)|divide(n2,n3)|power(#1,const_2)|multiply(#0,#2)
geometry
? % of 360 = 93.6
"? % of 360 = 93.6 or , ? = 93.6 Γ— 100 / 360 = 26 answer c"
a ) 277 , b ) 36 , c ) 26 , d ) 72 , e ) none of these
c
divide(multiply(93.6, const_100), 360)
multiply(n1,const_100)|divide(#0,n0)|
gain
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 450 , what were her original savings ?
"if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 450 . so 1 / 4 of her savings is $ 450 . so her original savings are 4 times $ 450 = $ 1800 correct answer d"
a ) $ 1500 , b ) $ 1600 , c ) $ 1700 , d ) $ 1800 , e ) $ 1900
d
divide(450, subtract(const_1, divide(3, 4)))
divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|
general
a man purchased 1 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ?
"10 * 150 = 1500 1 * 100 + 5 * 150 = 850 1500 – 850 = 650 answer : e"
a ) 278 , b ) 277 , c ) 278 , d ) 450 , e ) 650
e
subtract(multiply(const_10, 150), add(multiply(1, 100), multiply(5, 150)))
multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)|
general
rice weighing 29 / 4 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces )
"29 / 4 Γ· 4 = 29 / 16 pounds in each container 29 / 16 pounds * 16 ounces / pound = 29 ounces in each container the answer is c ."
a ) 21 , b ) 25 , c ) 29 , d ) 33 , e ) 37
c
divide(multiply(divide(29, 4), 16), 4)
divide(n0,n1)|multiply(n4,#0)|divide(#1,n2)|
general
the speed at which a man can row a boat in still water is 15 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 90 meters ?
"the speed of the boat downstream = 15 + 3 = 18 km / hr 18 km / hr * 5 / 18 = 5 m / s the time taken to cover 90 meters = 90 / 5 = 18 seconds . the answer is d ."
a ) 12 , b ) 16 , c ) 17 , d ) 18 , e ) 20
d
divide(90, multiply(add(15, 3), const_0_2778))
add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 9 ] * [ 10 ] equivalent to ?
[ 9 ] * [ 10 ] = 27 * 5 = 135 = 3 * 45 = [ 45 ] the answer is e .
a ) [ 25 ] , b ) [ 135 ] , c ) [ 30 ] , d ) [ 90 ] , e ) [ 45 ]
e
divide(multiply(divide(10, 2), multiply(3, 9)), const_3)
divide(n4,n2)|multiply(n0,n3)|multiply(#0,#1)|divide(#2,const_3)
general
a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability q that a + b is prime ?
"total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . to...
a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9
c
multiply(add(const_12, const_3), power(divide(1, 6), const_2))
add(const_12,const_3)|divide(n0,n1)|power(#1,const_2)|multiply(#0,#2)|
general
arun purchased 30 kg of wheat at the rate of rs . 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg . he mixed the two and sold the mixture . approximately what price per kg should be sell the mixture to make 20 % profit ?
"c . p . of 50 kg wheat = ( 30 * 11.50 + 20 * 14.25 ) = rs . 630 . s . p . of 50 kg wheat = 120 % of rs . 630 = 120 / 100 * 630 = rs . 756 s . p . per kg = 756 / 50 = rs . 15.12 answer : a"
a ) 15.12 , b ) 16.33 , c ) 16.35 , d ) 16.3 , e ) 16.32
a
divide(add(add(multiply(30, 11.50), multiply(20, 14.25)), multiply(divide(add(multiply(30, 11.50), multiply(20, 14.25)), const_100), 20)), add(30, 20))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,const_100)|multiply(n4,#4)|add(#3,#5)|divide(#6,#0)|
gain
double of quarter of 2 percent written as a decimal is :
"explanation : solution : ( 2 ) * ( 1 / 4 ) * 2 % = 2 * ( 1 / 4 * 2 / 100 ) = 0.01 . answer : a"
a ) 0.01 , b ) 0.0005 , c ) 0.25 , d ) 0.005 , e ) none of these
a
multiply(divide(divide(2, const_100), 2), const_2)
divide(n0,const_100)|divide(#0,n0)|multiply(#1,const_2)|
gain
a boat covers a certain distance downstream in 1 hour , while it comes back in 11 ⁄ 2 hours . if the speed of the stream be 3 kmph , what is the speed of the boat in still water ?
"explanation : let the speed of the water in still water = x given that speed of the stream = 3 kmph speed downstream = ( x + 3 ) kmph speed upstream = ( x - 3 ) kmph he travels a certain distance downstream in 1 hour and come back in 11 ⁄ 2 hour . ie , distance travelled downstream in 1 hour = distance travelled upstr...
a ) 14 kmph , b ) 15 kmph , c ) 13 kmph , d ) 12 kmph , e ) none of these
b
divide(add(multiply(divide(const_3, const_2), const_3.0), 2), subtract(divide(const_3, const_2), 1))
divide(const_3,const_2)|multiply(const_3.0,#0)|subtract(#0,n0)|add(const_3.0,#1)|divide(#3,#2)|
physics
a certain bag contains 100 balls Γ’ € ” 50 white , 30 green , 8 yellow , 9 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ?
"according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 50 + 30 + 8 ) / 100 = 88 / 100 = 0.88 . answer is d"
a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.88 , e ) 0.5
d
divide(subtract(100, add(9, 3)), 100)
add(n4,n5)|subtract(n0,#0)|divide(#1,n0)|
other
if the wheel is 9 cm then the number of revolutions to cover a distance of 1056 cm is ?
"2 * 22 / 7 * 9 * x = 1056 = > x = 18.7 answer : d"
a ) 18 , b ) 26 , c ) 14 , d ) 18.7 , e ) 91
d
divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 9))
multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)|
physics
kanul spent $ 80000 in buying raw materials , $ 30000 in buying machinery and 20 % of the total amount he had as cash with him . what was the total amount ?
"let the total amount be x then , ( 100 - 20 ) % of x = 80000 + 30000 80 % of x = 110000 80 x / 100 = 110000 x = $ 1100000 / 8 x = $ 137500 answer is b"
a ) $ 135656 , b ) $ 137500 , c ) $ 134446 , d ) $ 123265 , e ) $ 124564
b
divide(add(80000, 30000), subtract(const_1, divide(20, const_100)))
add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2)|
gain
the product of two numbers is 4107 . if the h . c . f . of these numbers is 37 , then the greater number is :
"explanation let the numbers be 37 a and 37 b . then , 37 a x 37 b = 4107 ab = 3 . now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 x 1 , 37 x 3 ) i . e . , ( 37 , 111 ) . greater number = 111 . answer c"
a ) 101 , b ) 107 , c ) 111 , d ) 185 , e ) none
c
sqrt(add(power(sqrt(subtract(37, multiply(const_2, 4107))), const_2), multiply(const_4, 4107)))
multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)|
general
if the operation ΓΈ is defined for all positive integers x and w by x ΓΈ w = ( 2 ^ x ) / ( 2 ^ w ) then ( 3 ΓΈ 1 ) ΓΈ 2 = ?
"3 ΓΈ 1 = 2 ^ 3 / 2 ^ 1 = 4 4 ΓΈ 2 = 2 ^ 4 / 2 ^ 2 = 4 the answer is b ."
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32
b
divide(power(2, divide(power(2, 3), power(2, 2))), power(2, 2))
power(n0,n2)|power(n0,n0)|power(n0,n4)|divide(#0,#1)|power(n0,#3)|divide(#4,#2)|
general
if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 63 , what is the value of x ?
"am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 63 x = 61 answer : d"
a ) 60 , b ) 62 , c ) 64 , d ) 61 , e ) 66
d
subtract(multiply(4, const_2), multiply(2, const_2))
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
general
a movie buff owns movies on dvd and on blu - ray in a ratio of 9 : 2 . if she buys 2 more blu - ray movies , that ratio would change to 17 : 4 . if she owns movies on no other medium , what was the original number of movies in her library before the extra purchase ?
"if u can just keep an eye on the options 99 is the only multiple of 9 in options given . . so you can mark it wid in seconds . now coming to the process m ( d ) = 9 x and b ( d ) = 2 x now from the next line the new eqn becomes 9 x / ( 2 x + 2 ) = 17 / 4 solving it 36 x = 34 x + 34 x = 17 which means m ( d ) = 153 and...
a ) 22 , b ) 28 , c ) 77 , d ) 99 , e ) 187
e
multiply(divide(multiply(17, add(2, const_1)), subtract(multiply(4, 9), multiply(17, 2))), add(9, 2))
add(n0,n1)|add(n2,const_1)|multiply(n0,n4)|multiply(n1,n3)|multiply(n3,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(#0,#6)|
general
if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 5 strokes ?
left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 left after 3 rd stroke = 1 / 2 * 1 / 4 = 1 / 8 left after 4 th stroke = 1 / 2 * 1 / 8 = 1 / 16 left after 5 th stroke = 1 / 2 * 1 / 16 = 1 / 32 so removed = 1 - 1 / 32 = 31 / 32
a ) 31 / 32 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16
a
add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2))
divide(n0,n1)|divide(#0,n1)|add(#0,#1)|divide(#1,n1)|add(#2,#3)|divide(#3,n1)|add(#4,#5)|divide(#5,n1)|add(#6,#7)
physics
the ratio between the length and the breadth of a rectangular park is 4 : 1 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 min , then the area of the park ( in sq . m ) is ?
"perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 4 x metres and breadth = 1 x metres . then , 2 ( 4 x + 1 x ) = 1600 or x = 160 . length = 640 m and breadth = 160 m . area = ( 640 x 160 ) m 2 = 102400 m e"
a ) 1024545 m , b ) 134561 m , c ) 156787 m , d ) 15450 m , e ) 102400 m
e
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(4, 1)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(4, const_4.0)), const_2), 1))
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(const_4.0,#7)|rectangle_area(#7,#8)|
physics
the average weight of 9 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ?
"total weight increases = 9 Γ— 1.5 = 13.5 kg so the weight of new person = 65 + 13.5 = 78.5 kg answer c"
a ) 76 kg , b ) 77 kg , c ) 78.5 kg , d ) data inadequate , e ) none of these
c
add(65, multiply(9, 1.5))
multiply(n0,n1)|add(n2,#0)|
general
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1344 ?
"1344 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1200 answer : c"
a ) 1000 , b ) 2217 , c ) 1200 , d ) 2777 , e ) 2991
c
divide(1344, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1))
multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|
general
there are 100 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of manager to 98 % ?
"we have 99 managers and 1 director . that 1 director to compose 2 % of the total number of people , there must be 50 people in the room , hence 50 managers must leave . answer : c ."
a ) 1 , b ) 2 , c ) 50 , d ) 49 , e ) 97
c
divide(subtract(multiply(100, divide(99, const_100)), multiply(100, divide(98, const_100))), subtract(const_1, divide(98, const_100)))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)|
gain
a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , b receives :
"let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 35000 x 17 / 50 ) = rs . 11,900 . e"
a ) s . 11,000 , b ) s . 11,300 , c ) s . 11,500 , d ) s . 11,700 , e ) s . 11,900
e
subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1)
add(n1,n2)|add(const_3,const_4)|multiply(n2,const_10)|multiply(n2,#1)|subtract(#2,n2)|subtract(#4,#0)|divide(#5,const_3)|add(#0,#6)|divide(#7,#2)|multiply(#8,#3)|divide(#9,const_1000)|floor(#10)|subtract(#11,const_1)|
general
how many 3 digit positive integers e exist that when divided by 7 leave a remainder of 5 ?
"minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leave...
a ) 128 , b ) 142 , c ) 143 , d ) 141 , e ) 129
e
divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7)
add(const_10,const_4)|multiply(const_10,const_100)|multiply(n1,#0)|subtract(#1,const_1)|add(n2,#2)|subtract(#3,#4)|divide(#5,n1)|
general
6.2 is what percent of 1000 ?
"10 % of 1000 is 100 . 6.2 is way less than 100 ( 10 % of 1000 ) . so it can not be 62 % of 1000 . eliminate option a 1 % of 1000 is 10 . 6.2 is less than 10 . so it cant be 6.2 % . eliminate option b ( 0.5 % ) i . e . half of 1 % of 1000 is 5 . but 6.2 is greater than 5 . . so it has to be greater than 0.5 % . . . hen...
a ) 62 % , b ) 6.2 % , c ) 0.62 % , d ) 0.062 % , e ) 0.0062 %
c
multiply(divide(6.2, 1000), const_100)
divide(n0,n1)|multiply(#0,const_100)|
gain
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 12 cm . what is the length of the wire ?
"volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . Ο€ ( 12 ) ^ 2 * h = ( 4 / 3 ) Ο€ ( 12 ) ^ 3 = > h = 16 cm answer : c"
a ) 6 cm , b ) 14 cm , c ) 16 cm , d ) 23 cm , e ) 29 cm
c
divide(multiply(const_4, divide(power(12, const_3), power(12, const_2))), const_3)
power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)|
physics
a van takes 6 hours to cover a distance of 252 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ?
"time = 6 distence = 252 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 252 / 9 = 28 kmph a"
a ) 28 kmph , b ) 60 kmph , c ) 70 kmph , d ) 80 kmph , e ) 90 kmph
a
divide(252, multiply(divide(3, 2), 6))
divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)|
physics
the ages of two persons differ by 28 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively
"explanation : let their ages be x and ( x + 28 ) years . 5 ( x - 5 ) = ( x + 28 - 5 ) or 4 x = 48 or x = 12 . their present ages are 40 years and 12 years option a"
a ) 40,12 , b ) 20,10 , c ) 25,15 , d ) 30,10 , e ) none of these
a
subtract(add(divide(multiply(28, 5), subtract(5, const_1)), 5), 28)
multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)|
general
3 x + y = 21 , and x + 3 y = 1 . find the value of 2 x + 2 y
"add these two equations 4 x + 4 y = 22 divide by 2 ( to get 2 x + 2 y ) answer will be d . 11"
a ) 20 , b ) 18 , c ) 16 , d ) 11 , e ) 5
d
divide(add(21, 1), 2)
add(n1,n3)|divide(#0,n5)|
general
x can finish a work in 20 days . y can finish the same work in 15 days . yworked for 9 days and left the job . how many days does x alone need to finish the remaining work ?
"work done by x in 1 day = 1 / 20 work done by y in 1 day = 1 / 15 work done by y in 9 days = 9 / 15 = 3 / 5 remaining work = 1 – 3 / 5 = 2 / 5 number of days in which x can finish the remaining work = ( 2 / 5 ) / ( 1 / 20 ) = 8 d"
a ) 3 , b ) 5 , c ) 6 , d ) 8 , e ) 9
d
divide(subtract(const_1, multiply(9, divide(const_1, 15))), divide(const_1, 20))
divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)|
physics
thomas worked in the factory @ 4550 / - per week . owner told him to give the amount after one month . then how much amount will be paid to thomas ?
total salary paid to thomas in one week : 4550 / - salary for one day : 4550 / 7 = 650 / - total for one month = 650 / - * 30 days = 19500 / - answer is option ( d )
a ) 9500 , b ) 17500 , c ) 18200 , d ) 19500 , e ) 6750
d
add(add(multiply(multiply(const_100, const_10), const_10), multiply(4550, const_2)), multiply(const_100, const_4))
multiply(const_10,const_100)|multiply(n0,const_2)|multiply(const_100,const_4)|multiply(#0,const_10)|add(#3,#1)|add(#4,#2)
general
in the first 10 overs of a cricket game , the run rate was only 3.6 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ?
"10 overs - run rate = 3.6 runs scored in first 10 overs = 36 remaining overs 40 total runs to be scored = 282 32 runs already scored 282 - 36 = 246 246 runs to be scored in 40 overs let required runrate be x 40 * x = 246 x = 246 / 40 x = 6.15 the required runrate is 6.15 answer : c"
a ) 6.25 , b ) 6.5 , c ) 6.15 , d ) 7 , e ) 8
c
divide(subtract(282, multiply(10, 3.6)), 40)
multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|
gain
in a tree , 3 / 5 of the birds are parrots while the rest are toucans . if 1 / 3 of the parrots are female and 3 / 4 of the toucans are female , what fraction of the birds in the tree are male ?
the fraction of birds that are male parrots is ( 2 / 3 ) ( 3 / 5 ) = 2 / 5 . the fraction of birds that are male toucans is ( 1 / 4 ) ( 2 / 5 ) = 1 / 10 . the total fraction of male birds is 2 / 5 + 1 / 10 = 1 / 2 . the answer is a .
a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 5 , d ) 3 / 10 , e ) 7 / 10
a
add(multiply(divide(3, 5), subtract(const_1, divide(1, 3))), multiply(divide(const_2, 5), subtract(const_1, divide(3, 4))))
divide(n0,n1)|divide(n2,n0)|divide(const_2,n1)|divide(n0,n5)|subtract(const_1,#1)|subtract(const_1,#3)|multiply(#0,#4)|multiply(#2,#5)|add(#6,#7)
general
working together at their respective constant rates , machine a and machine b can produce 1,800 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 1,800 unit total would it produce ?
"1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 1800 units in 24 hrs , so it can produce 1800 * 8 / 24 = 600 units is 8 hrs . 600 is 30 % of 1800 . c is the answer"
a ) 25 , b ) 33 , c ) 30 , d ) 67 , e ) 75
c
multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100)
multiply(const_4,const_4)|multiply(n1,n2)|divide(#1,const_100)|multiply(#0,const_100)|add(n1,#2)|divide(#3,n1)|divide(#3,#4)|subtract(#5,#6)|multiply(n1,#7)|divide(#8,#3)|multiply(#9,const_100)|
gain
a coin is tossed 6 times . what is the probability of getting exactly 5 heads ?
"the number of possible outcomes is 2 ^ 6 = 64 there are 6 ways to get exactly 5 heads . p ( exactly 5 heads ) = 6 / 64 = 3 / 32 the answer is d ."
a ) 3 / 8 , b ) 3 / 16 , c ) 5 / 16 , d ) 3 / 32 , e ) 7 / 64
d
multiply(power(divide(const_1, const_2), 5), multiply(choose(6, 5), divide(const_1, const_2)))
choose(n0,n1)|divide(const_1,const_2)|multiply(#0,#1)|power(#1,n1)|multiply(#2,#3)|
probability
at a supermarket , john spent 1 / 4 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 6 on bakery products . if he spent the remaining $ 6 on candy , how much did john spend at the supermarket ?
"let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 4 of his money on fresh fruits and vegetables , or ( 1 / 4 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 6 on bakery products , or ( 1 / 6 ) t . w...
a ) $ 24 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 180
a
divide(6, subtract(1, add(add(divide(1, 6), divide(1, 3)), divide(1, 4))))
divide(n0,n5)|divide(n0,n3)|divide(n0,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)|
general
a and b are two multiples of 14 , and q is the set of consecutive integers between a and b , inclusive . if q contains 13 multiples of 14 , how many multiples of 7 are there in q ?
halfway between the multiples of 14 , there will be another multiple of 7 . the total number of multiples of 7 is 13 + 12 = 25 . the answer is c .
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
c
subtract(multiply(13, const_2), const_1)
multiply(n1,const_2)|subtract(#0,const_1)
physics
the population of a bacteria colony doubles every day . if it was started 6 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ?
"2 ^ 6 ( 2 ) = 2 ^ 7 = 128 the answer is e ."
a ) 512 , b ) 768 , c ) 1024 , d ) 2048 , e ) 128
e
subtract(power(2, add(6, const_1)), const_1)
add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|
physics
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , the number of valid votes that the other candidate got , was :
"b number of valid votes = 80 % of 7500 = 6000 . valid votes polled by other candidate = 45 % of 6000 = ( 45 / 100 x 6000 ) = 2700 ."
a ) 2800 , b ) 2700 , c ) 2900 , d ) 2200 , e ) 2300
b
multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500)
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)|
gain
the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 48.0 billion from $ 85.0 billion , a year ago . by what percent did the revenue fall ?
$ 85 - $ 48 = 37 $ ( 37 / 85 ) * 100 = 43.5 % answer : c
a ) 20.8 , b ) 30.4 , c ) 43.5 , d ) 32.5 , e ) 33.8
c
multiply(divide(subtract(85, 48), 85), const_100)
subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)
gain
10 ^ 23 - 7 is divided by 6 , remainder is ?
anything power by 10 we got 10 and some zeros after that . ( i . e , 1000000000000000 . . . . ) on that , we minus 7 means , we must get last two digits is 93 . and 93 is divided by 6 , we get reminder 3 . answer : a
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
subtract(subtract(subtract(23, 10), 6), const_4)
subtract(n1,n0)|subtract(#0,n3)|subtract(#1,const_4)
general
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 110 marks , the no of questions he attempts correctly is :
"explanation : let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 4 x – ( 60 – x ) = 110 = > 5 x = 170 = > x = 34 answer : a"
a ) 34 , b ) 38 , c ) 40 , d ) 42 , e ) 44
a
divide(add(110, 60), add(4, 1))
add(n2,n3)|add(n0,n1)|divide(#0,#1)|
physics
the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 3.6 miles in the sixth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ?
"( 3.6 miles / 8 minutes ) * 60 minutes / hour = 27 mph let x be the original speed . x - 5 ( 3 ) = 27 x = 42 mph the answer is d ."
a ) 30 , b ) 34 , c ) 38 , d ) 42 , e ) 46
d
add(add(add(add(divide(3.6, divide(8, const_60)), 3), 3), 3), 3)
divide(n1,const_60)|divide(n2,#0)|add(n0,#1)|add(n0,#2)|add(n0,#3)|add(n0,#4)|
general
in an office , 20 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ?
"( 10 / 100 ) workers = 16 = > number of workers = 160 ( 20 / 100 ) * workers = x + 16 = > x = 32 answer a"
a ) 32 , b ) 64 , c ) 50 , d ) 144 , e ) 160
a
divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4))
divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)|
gain
when a student joe , weighing 45 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left an...
"after two persons leave the group the average remains the same . that means the weight of the two persons = 45 + 30 = 75 so , the average the two persons = 37.5 that gives the answer 45 - 37.5 = 7.5 answer a"
a ) 7.5 kg , b ) 11 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg
a
subtract(45, divide(subtract(add(multiply(30, subtract(45, add(30, 1))), 45), multiply(subtract(subtract(45, add(30, 1)), 1), 30)), const_2))
add(n1,n2)|subtract(n0,#0)|multiply(n1,#1)|subtract(#1,n2)|add(n0,#2)|multiply(n1,#3)|subtract(#4,#5)|divide(#6,const_2)|subtract(n0,#7)|
general
what is the probability of getting a sum 9 from two throws of dice
"explanation : total number of cases = 6 * 6 = 36 favoured cases = [ ( 3,6 ) , ( 4,5 ) , ( 6,3 ) , ( 5,4 ) ] = 4 so probability = 4 / 36 = 1 / 9 answer : b"
a ) 1 / 3 , b ) 1 / 9 , c ) 1 / 12 , d ) 2 / 9 , e ) none of these
b
divide(const_2, choose(add(const_3, const_3), const_3))
add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)|
probability
if pintu is coded as 79523 in a certain code language , how would you code buntu in the same language ?
"1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . buntu is 23523 . . . answer : b"
a ) 13426 , b ) 23523 , c ) 13436 , d ) 14263 , e ) 15263
b
divide(79523, add(const_3, const_3))
add(const_3,const_3)|divide(n0,#0)|
general
if neha is 10 both sonali and priyanka is 15 both sadaf and tanu is 10 . how much is prinka by the same system ?
c prinka is 10 , in a system that awards five for each vowel .
a ) 8 , b ) 11 , c ) 10 , d ) 15 , e ) 6
c
multiply(subtract(15, 10), const_2)
subtract(n1,n0)|multiply(#0,const_2)
other
if ( a + b ) = 14 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ?
"given a + b = 14 = > a = 14 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 14 - b + 3 - c = > 17 - ( b + c ) = > 17 - 9 = 8 . option b . . ."
a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 .
b
subtract(add(14, 3), 9)
add(n0,n2)|subtract(#0,n1)|
general
during a sale , the price of a pair of shoes is marked down 13 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ?
"assume the price = 100 price during sale = 87 price after sale = 100 percent increase = 13 / 87 * 100 = 15 % approx . correct option : d"
a ) 9 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 90 %
d
divide(multiply(13, const_100), subtract(const_100, 13))
multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)|
gain
machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 4200 envelopes .
"you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b a...
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 60 / 7
b
divide(4200, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3))))
divide(n0,n2)|divide(n3,n4)|divide(n0,n1)|subtract(#1,#2)|subtract(#0,#3)|divide(n5,#4)|
physics
the average weight of 8 persons increases by 4.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ?
"total weight increased = ( 8 x 4.5 ) kg = 36 kg . weight of new person = ( 65 + 36 ) kg = 101 kg . answer : e"
a ) 75 kg , b ) 65 kg , c ) 55 kg , d ) 85 kg , e ) 101 kg
e
add(65, multiply(8, 4.5))
multiply(n0,n1)|add(n2,#0)|
general
if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 60 , what is the value of c Γ’ Λ† ’ a ?
"the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 120 - - 2 subtracting 1 from 2 we have c - a = 30 ; answer : b"
a ) 25 , b ) 30 , c ) 90 , d ) 140 , e ) it can not be determined from the information given
b
subtract(multiply(60, const_2), multiply(45, const_2))
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
general
the lcm and hcf of two numbers are 100000 and 10000 respectively . find the larger of the two numbers if their sum is 110000 .
"there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 100000 * 10000 = 1000000000 so the answer which we are looking for has to be a factor of 1000000000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 1000000000 . and then take the hig...
a ) 100000 , b ) 46 , c ) 577 , d ) 577 , e ) 767
a
divide(multiply(100000, 10000), 110000)
multiply(n0,n1)|divide(#0,n2)|
physics
when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum r of all possible 2 - digit remainders for x / y ?
"ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . r = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b"
a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784
b
divide(59.32, subtract(2, floor(2)))
floor(n1)|subtract(n1,#0)|divide(n0,#1)|
general
a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 5 years ago ?
"let ' s say age of richard isr age of david isd age of scott iss now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6...
a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 16
a
subtract(add(6, 8), 5)
add(n1,n2)|subtract(#0,n4)|
general
car dealership x , the total profit from sales increased by 15 percent over the previous years , while the number of cars sold decreased by 15 percent over the previous years . approximately what was the average percent increase in profit per car over the previous year .
"let profit be x and cars sold be y originally profit / car = x / y now it is 1.15 x / 0.85 y = 115 / 85 ( x / y ) increase in profit per car = ( ( 30 / 85 ) ( x / y ) ) / ( x / y ) * 100 = 35.29 % answer = d"
a ) 18 % , b ) 20 % , c ) 22 % , d ) 35.29 % , e ) 55 %
d
multiply(divide(subtract(divide(add(const_100, 15), subtract(const_10, const_3)), divide(const_100, const_10)), divide(const_100, const_10)), const_100)
add(n0,const_100)|divide(const_100,const_10)|subtract(const_10,const_3)|divide(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)|
general
jill has 42 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ?
let the number of each size of jar = wthen 1 / 4 w + 1 / 2 w + w = 42 1 3 / 4 w = 42 w = 24 the total number of jars = 3 w = 72 answer : d
a ) 3 , b ) 6 , c ) 9 , d ) 72 , e ) 14
d
multiply(divide(42, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3)
divide(const_1,const_2)|add(#0,const_0_25)|add(#1,const_1)|divide(n0,#2)|multiply(#3,const_3)|
general
a sum of money becomes 7 / 6 of itself in 4 years at a certain rate of simple interest . the rate per annum is ?
let sum = x . then , amount = 7 x / 6 s . i . = 7 x / 6 - x = x / 6 ; time = 4 years . rate = ( 100 * x ) / ( x * 6 * 4 ) = 25 / 6 % . answer : a
a ) 25 / 6 , b ) 5 5 / 7 , c ) 5 5 / 2 , d ) 5 5 / 8 , e ) 5 5 / 1
a
multiply(divide(subtract(divide(7, 6), const_1), 4), const_100)
divide(n0,n1)|subtract(#0,const_1)|divide(#1,n2)|multiply(#2,const_100)
gain
of the 3 - digit integers greater than 700 , how many have distinct digits ?
three digit number can have only following 3 patterns : a . all digits are distinct ; b . two digits are alike and third is different ; c . all three digits are alike . we need to calculate b . b = total - a - c total numbers from 700 to 999 = 299 ( 3 - digit numbers greater than 700 ) ; a . all digits are distinct = 3...
a ) 216 , b ) 82 , c ) 80 , d ) 45 , e ) 36
a
multiply(multiply(3, subtract(const_10, const_1)), subtract(const_10, const_2))
subtract(const_10,const_1)|subtract(const_10,const_2)|multiply(n0,#0)|multiply(#2,#1)
general
a can finish a work in 9 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ?
"b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 9 * 1 / 3 ) = 3 days . correct option : a"
a ) 3 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these
a
divide(multiply(multiply(divide(const_1, 15), 10), 9), const_2)
divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)|
physics
let f ( x ) = x ^ 2 + bx + c . if f ( 6 ) = 0 and f ( - 2 ) = 0 , then b + c =
"f ( x ) = x ^ 2 + bx + c . if f ( 6 ) = 0 and f ( - 2 ) = 0 , then b + c = f ( 6 ) = 0 = 36 + 6 b + c - - - taking 36 to the other side - > 6 b + c = - 36 f ( - 2 ) = 0 = 4 - 2 b + c - - - taking - 2 b + c to the other side - > 2 b - c = 4 when we add these 2 equations , we get 8 b = - 32 - - - > b = - 4 and while sub...
a ) 18 , b ) 15 , c ) - 16 , d ) - 21 , e ) - 24
c
negate(divide(subtract(power(2, 2), 6), add(2, 6)))
add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)|
general
a wire can be bent in the form of a circle of radius 56 cm . if it is bent in the form of a square , then its area will be
"length of wire = 2 pier = 2 * ( 22 / 7 ) * 56 = 352 cm side of the square = 352 / 4 = 88 cm area of the square = 88 * 88 = 7744 sq cm answer : a"
a ) 7744 , b ) 8844 , c ) 5544 , d ) 4444 , e ) 3444
a
power(divide(circumface(56), const_4), const_2)
circumface(n0)|divide(#0,const_4)|power(#1,const_2)|
geometry
if 40 % of a number is equal to two - third of another number , what is the ratio of first number to the second number ?
"let 40 % of a = 2 / 3 b then 40 a / 100 = 2 b / 3 2 a / 5 = 2 b / 3 a / b = 5 / 3 a : b = 5 : 3 answer is d"
a ) 2 : 5 , b ) 1 : 4 , c ) 3 : 7 , d ) 5 : 3 , e ) 2 : 3
d
divide(divide(const_1, const_4), divide(40, const_100))
divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)|
general