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pgilliar
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MNLP_M2_rag_dataset

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Rationale
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if the wheel is 14 cm then the number of revolutions to cover a distance of 3520 cm is ?
"2 * 22 / 7 * 14 * x = 3520 = > x = 40 answer : e"
a ) 22 , b ) 28 , c ) 17 , d ) 12 , e ) 40
e
divide(3520, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14))
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
E
in a 280 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is :
"b runs 56 m in 7 sec . = > b runs 280 m in 7 / 56 * 280 = 35 seconds since a beats b by 7 seconds , a runs 280 m in ( 35 - 7 ) = 28 seconds hence , a ' s time over the course = 28 seconds answer : e"
a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds
e
subtract(multiply(divide(7, 56), 280), 7)
divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)|
physics
E
the mean of 50 observations was 32 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is
"sol . therefore correct sum = ( 32 Γ— 50 + 48 – 23 ) = 1625 . therefore correct mean = 1625 / 50 = 32.5 . answer b"
a ) 35.2 , b ) 32.5 , c ) 36.5 , d ) 39.1 , e ) none
b
divide(add(multiply(32, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
B
a card board of 34 * 14 has to be attached to a wooden box and a total of 35 pins are to be used on the each side of the cardbox . find the total number of pins used .
( 35 * 4 ) - 4 = 136 answer : b
a ) 135 , b ) 136 , c ) 137 , d ) 138 , e ) 139
b
multiply(35, const_4)
multiply(n2,const_4)
general
B
the cash realised on selling a 14 % stock is rs . 106.25 , brokerage being 1 / 4 % is
explanation : cash realised = rs . ( 106.25 - 0.25 ) = rs . 106 . answer : b
a ) 123 , b ) 106 , c ) 100 , d ) 156 , e ) 240
b
subtract(106.25, divide(1, 4))
divide(n2,n3)|subtract(n1,#0)
general
B
a car travels uphill at 30 km / hr and downhill at 70 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ?
avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 60 = 5 / 7 ; avg speed = 150 / ( 10 / 3 + 5 / 7 ) = 37 kmph answer : e
a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 37 kmph
e
divide(add(100, 50), add(divide(100, 30), divide(50, 70)))
add(n2,n3)|divide(n2,n0)|divide(n3,n1)|add(#1,#2)|divide(#0,#3)
general
E
8900 Γ· 6 Γ· 4 = ?
explanation : given exp . 8900 * 1 / 6 * 1 / 4 = 370.833 answer : d
a ) 349 , b ) 541.75 , c ) 224.37 , d ) 370.833 , e ) none of these
d
divide(divide(8900, 6), 4)
divide(n0,n1)|divide(#0,n2)
general
D
a person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 30 hour to row to a place and come back , how far is the place ?
"speed of down stream = 10 + 2 = 12 kmph speed of upstream = 10 - 2 = 8 kmph let the required distance be xkm x / 12 + x / 8 = 30 2 x + 3 x = 720 x = 144 km answer is a"
a ) 144 km , b ) 30 km , c ) 48 km , d ) 12 km , e ) 15 km
a
divide(multiply(multiply(subtract(10, 2), add(10, 2)), 30), add(subtract(10, 2), add(10, 2)))
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n2,#3)|divide(#4,#2)|
physics
A
the batting average of a particular batsman is 58 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score .
"explanation : total runs scored by the batsman = 58 * 46 = 2668 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2668 – 2552 = 116 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 116 2 x = 266 x = 133 runs answer d"
a ) 179 , b ) 208 , c ) 210 , d ) 133 , e ) 229
d
divide(add(150, subtract(multiply(58, 46), multiply(58, subtract(46, const_2)))), const_2)
multiply(n0,n1)|subtract(n1,const_2)|multiply(n3,#1)|subtract(#0,#2)|add(n2,#3)|divide(#4,const_2)|
general
D
the area of a rectangular field is equal to 200 square meters . its perimeter is equal to 60 meters . find the width of this rectangle .
"l * w = 200 : area , l is the length and w is the width . 2 l + 2 w = 60 : perimeter l = 30 - w : solve for l ( 30 - w ) * w = 200 : substitute in the area equation w = 10 and l = 20 : correct answer b"
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25
b
divide(subtract(divide(60, const_2), sqrt(subtract(multiply(divide(60, const_2), divide(60, const_2)), multiply(const_4, 200)))), const_2)
divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)|
geometry
B
in what time will a train 100 m long cross an electric pole , it its speed be 126 km / hr ?
"speed = 126 * 5 / 18 = 35 m / sec time taken = 100 / 35 = 2.85 sec . answer : d"
a ) 2.5 , b ) 2.9 , c ) 2.4 , d ) 2.85 , e ) 2.1
d
divide(100, multiply(126, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
D
what is the units digit of the expression 14 ^ 7 βˆ’ 17 ^ 4 ?
i think answer on this one should be d too . since we know that 14 ^ 7 > 17 ^ 4 , as will said one should always check if the number is positive .
a ) 0 , b ) 3 , c ) 4 , d ) 8 , e ) 6
d
add(add(const_4, const_3), const_2)
add(const_3,const_4)|add(#0,const_2)|
general
D
find two integers , neither of which ends in a zero , and whose product is exactly 10 , 000,000
"10 , 000,000 = 10 ^ 7 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) = ( 2 ^ 7 ) x ( 5 ^ 7 ) = 128 x 78125 so the numbers are 128 and 78,125 answer : e"
a ) 64 and 15,625 , b ) 60 and 15,625 , c ) 64 and 15,620 , d ) 64 and 15,635 , e ) 128 and 78,125
e
divide(multiply(multiply(const_100, const_100), const_100), divide(multiply(multiply(const_100, const_100), const_100), subtract(multiply(const_3, const_12), const_4)))
multiply(const_100,const_100)|multiply(const_12,const_3)|multiply(#0,const_100)|subtract(#1,const_4)|divide(#2,#3)|divide(#2,#4)|
general
E
a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 130 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 120 = 370 m . time taken = 370 / 10 = 37 sec . answer : c"
a ) 28 sec , b ) 16 sec , c ) 37 sec , d ) 18 sec , e ) 17 sec
c
divide(add(240, 130), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))
add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|
general
C
if the price of a certain computer increased 30 percent from d dollars to 338 dollars , then 2 d =
"before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 338 ( given ) i . e . d = 338 / 1.3 = $ 260 i . e . 2 d = 2 * 260 = 520 answer : option b"
a ) 540 , b ) 520 , c ) 619 , d ) 649 , e ) 700
b
multiply(divide(338, divide(add(const_100, 30), const_100)), 2)
add(n0,const_100)|divide(#0,const_100)|divide(n1,#1)|multiply(n2,#2)|
general
B
what is the square root of 400 , divided by 2 ?
"square root is a number times itself square root of 400 = 20 , 20 / 2 = 10 ( e ) 1"
a ) 9 , b ) 36 , c ) 122 , d ) 6 , e ) 10
e
divide(sqrt(400), 2)
sqrt(n0)|divide(#0,n1)|
other
E
the jogging track in a sports complex is 561 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ?
clearly , the two will meet when they are 561 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 561 m apart , they take ( 100 / 825 * 561 / 1000 ) hrs = ( 561 / 8250 * 60 ) min = 4.08 min . answer : c
a ) 5.29 min , b ) 5.28 min , c ) 4.08 min , d ) 9.28 min , e ) 5.988 min
c
multiply(divide(divide(561, const_1000), add(4.5, 3.75)), const_60)
add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_60)
general
C
in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 40 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ?
"the percent of tagged fish in the second catch is 2 / 40 * 100 = 5 % . we are told that 5 % approximates the percent of tagged fish in the pond . since there are 40 tagged fish , then we have 0.05 x = 40 - - > x = 800 . answer : b ."
a ) 400 , b ) 800 , c ) 1250 , d ) 2500 , e ) 10 000
b
divide(40, divide(2, 40))
divide(n2,n1)|divide(n0,#0)|
gain
B
how long will take a leak at the bottom of a tank to empty it if it will take 3 hours to fill it without the leak , but due to the leak it will take one additional hour to be filled ?
part filled without leak in 1 hour = 1 / 3 part filled with leak in 1 hour = 1 / 4 work done by leak in 1 hour = 1 / 3 Γ’ Λ† ’ 1 / 4 = 12 hours answer : a
a ) 12 hours , b ) 111 hours , c ) 15 hours , d ) 14 hours , e ) 11 hours
a
inverse(subtract(inverse(3), inverse(add(3, const_1))))
add(n0,const_1)|inverse(n0)|inverse(#0)|subtract(#1,#2)|inverse(#3)
physics
A
an alloy of copper and zinc contains copper and zinc in the ratio 3 : 5 . another alloy of copper and zinc contains copper and zinc in the ratio 6 : 2 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ?
"let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 3 x / 8 , and fraction of zinc in alloy _ 1 = 5 x / 8 . similarly , fraction of copper in alloy _ 2 = 6 y / 8 , and fraction of zinc in alloy _ 2 = 2 y / 8 . mixing them , we get copper = 3 x / 8 + 6 y / 8 ; zinc = 5 x / 8 + 2 y / 8 . so , 3 x + 6 y = 5 x + 2 y - > 2 x = 4 y - > x / y = 2 / 4 = 1 / 2 so , they must be mixed in the ratio 1 : 2 answer : a"
a ) 1 : 2 , b ) 2 : 2 , c ) 2 : 5 , d ) 2 : 6 , e ) 2 : 7
a
divide(subtract(divide(3, add(6, 2)), divide(6, 2)), subtract(divide(6, 2), divide(6, add(6, 2))))
add(n2,n3)|divide(n2,n3)|divide(n0,#0)|divide(n2,#0)|subtract(#2,#1)|subtract(#1,#3)|divide(#4,#5)|
general
A
in a factory , there are 50 % technicians and 50 % non - technicians . if the 50 % of the technicians and 50 % of non - technicians are permanent employees , then the percentage of workers who are temporary is ?
"total = 100 t = 50 nt = 50 50 * ( 50 / 100 ) = 25 50 * ( 50 / 100 ) = 25 25 + 25 = 50 = > 100 - 50 = 50 % answer : a"
a ) 50 % , b ) 57 % , c ) 52 % , d ) 22 % , e ) 42 %
a
subtract(const_100, add(multiply(50, divide(50, const_100)), multiply(divide(50, const_100), 50)))
divide(n1,const_100)|divide(n0,const_100)|multiply(n0,#0)|multiply(n1,#1)|add(#2,#3)|subtract(const_100,#4)|
gain
A
a certain no . when divided by 50 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
"explanation : 50 + 25 = 75 / 15 = 5 ( remainder ) c"
a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 9
c
subtract(subtract(subtract(50, 25), const_4), const_2)
subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)|
general
C
the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is ?
"let their ages be x years and ( x + 10 ) years then , ( x + 10 ) - 15 = 2 ( x - 15 ) x - 5 = 2 x - 30 x = 25 present age of the elder person = 25 + 10 = 35 years answer is a"
a ) 35 yr , b ) 25 yr , c ) 20 yr , d ) 30 yr , e ) 40 yr
a
add(subtract(10, subtract(10, add(const_3, const_2))), multiply(subtract(10, 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
noelle walks from point a to point b at an average speed of 6 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 7 kilometers per hour ?
"let ' s suppose that speed while returning was xkm / h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 / s 1 + s 2 7 = 2 * 6 * x / 6 + x x = 8.4 e is the answer"
a ) 6.75 , b ) 7 , c ) 7.25 , d ) 7.5 , e ) 8.4
e
divide(multiply(7, 6), subtract(multiply(const_2, 6), 7))
multiply(n0,n1)|multiply(n0,const_2)|subtract(#1,n1)|divide(#0,#2)|
physics
E
a semicircle has a radius of 9 . what is the approximate perimeter of the semicircle ?
"the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r / 2 + 2 r = pi * r + 2 r the perimeter is pi * 9 + 2 * 9 which is about 46 . the answer is d ."
a ) 16 , b ) 25 , c ) 33 , d ) 46 , e ) 58
d
add(divide(circumface(9), const_2), multiply(const_2, 9))
circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)|
geometry
D
there are 3 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ?
"there are 3 red chips and 2 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? 10 12 24 60 100 soln : total number of patterns is 5 ! since 3 red chips are identical and 2 blue ones are identical thus we have = 5 ! / ( 2 ! * 3 ! ) = 10 such different patterns a"
a ) 10 , b ) 12 , c ) 24 , d ) 60 , e ) 100
a
multiply(factorial(3), factorial(2))
factorial(n0)|factorial(n1)|multiply(#0,#1)|
general
A
0.0006688 / 0.0000150 x 19.85 = ?
"explanation : ? = 6688 / 150 x 19.85 = 45 x 20 = 900 answer : option b"
a ) 850 , b ) 900 , c ) 950 , d ) 1000 , e ) 2000
b
divide(0.0006688, 0.0000150)
divide(n0,n1)|
general
B
if rs . 578 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ?
"( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 578 * 2 / 17 = rs . 68 b ' s share = 578 * 3 / 17 = rs . 102 c ' s share = 578 * 12 / 17 = rs . 408 . answer : b"
a ) s . 300 , b ) s . 408 , c ) s . 389 , d ) s . 368 , e ) s . 323
b
divide(578, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1))
divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)|
general
B
sum of the squares of 3 no . ' s is 222 and the sum of their products taken two at a time is 131 . find the sum ?
"( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 222 + 2 * 131 a + b + c = √ 484 = 22 a"
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30
a
sqrt(add(multiply(131, const_2), 222))
multiply(n2,const_2)|add(n1,#0)|sqrt(#1)|
general
A
how many seconds does sandy take to cover a distance of 500 meters , if sandy runs at a speed of 18 km / hr ?
"18 km / hr = 18000 m / 3600 s = 5 m / s time = 500 / 5 = 100 seconds the answer is b ."
a ) 80 , b ) 100 , c ) 120 , d ) 140 , e ) 160
b
divide(500, multiply(18, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
B
( 0.96 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.96 ) ( power 2 ) + 0.096 + ( 0.1 ) ( power 2 ) is :
"given expression = ( 0.96 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.96 ) ( power 2 ) + ( 0.96 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.96 - 0.1 ) = 0.86 answer is c ."
a ) 0.68 , b ) 0.086 , c ) 0.86 , d ) 0.068 , e ) none of them
c
divide(subtract(power(0.96, 3), power(0.1, 3)), add(add(power(0.96, 2), 0.096), power(0.1, 2)))
power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)|
general
C
by selling an article for $ 110 , a person gains $ 10 . what is the gain % ?
"s . p . = $ 110 gain = $ 10 c . p . = 110 - 10 = 100 gain % = 10 / 100 * 100 % = 10 % answer is e"
a ) 25 % , b ) 30 % , c ) 50 % , d ) 20 % , e ) 10 %
e
divide(multiply(10, const_100), subtract(110, 10))
multiply(n1,const_100)|subtract(n0,n1)|divide(#0,#1)|
gain
E
if the charge of staying in a student youth hostel $ 18.00 / day for the first week , and $ 14.00 / day for each additional week , how much does it cost to stay for 23 days ?
"total number of days of stay = 23 charge of staying in first week = 18 * 7 = 126 $ charge of staying for additional days = ( 23 - 7 ) * 14 = 16 * 14 = 224 $ total charge = 126 + 224 = 350 $ answer b"
a ) $ 160 , b ) $ 350 , c ) $ 282 , d ) $ 274 , e ) $ 286
b
add(multiply(18.00, add(const_3, const_4)), multiply(14.00, subtract(23, add(const_3, const_4))))
add(const_3,const_4)|multiply(n0,#0)|subtract(n2,#0)|multiply(n1,#2)|add(#1,#3)|
general
B
a pet groomer has 7 animals to groom for the day ( 2 cats and 5 dogs ) . if she randomly selects 4 animals to groom before lunch , what is the probability she will finish all the cats before lunch ?
combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 4 animals from 7 animals = 7 c 4 = 7 ! / [ 4 ! ( 7 - 4 ) ! ] = 35 . to finish all 2 cats there must be 2 dogs , select 2 dogs from 5 = 5 c 2 = 10 . and , select 2 cats from 2 = 2 c 2 = 1 . 2 cats and 2 dogs = ( 1 ) ( 10 ) = 10 probability = ( number outcomes favorable ) / ( total number outcomes ) = 10 / 735 = 2 / 7 answer : b
a ) 3 / 7 , b ) 2 / 7 , c ) 12 / 13 , d ) 5 / 11 , e ) 4 / 7
b
divide(2, 7)
divide(n1,n0)
physics
B
the difference between a two - digit number and the number obtained by interchanging the digit is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
since the number is greater than the number obtained on reversing the digits , so the ten ' s digit is greater than the unit ' s digit . let the ten ' s and unit ' s digits be 2 x and x respectively . then , ( 10 * 2 x + x ) - ( 10 x + 2 x ) = 36 9 x = 36 x = 4 required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 . answer a
a ) 8 , b ) 15 , c ) 14 , d ) 12 , e ) 10
a
subtract(add(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2)), subtract(multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2), divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2))))
multiply(n2,const_10)|add(n1,#0)|subtract(#1,const_10)|subtract(#2,n2)|divide(n0,#3)|multiply(n2,#4)|add(#4,#5)|subtract(#5,#4)|subtract(#6,#7)
general
A
a trader bought a car at 30 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ?
"original price = 100 cp = 70 s = 70 * ( 140 / 100 ) = 98 100 - 112 = 2 % answer : e"
a ) 118 , b ) 110 , c ) 112 , d ) 113 , e ) 98
e
multiply(subtract(divide(divide(multiply(subtract(const_100, 30), add(const_100, 40)), const_100), const_100), const_1), const_100)
add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
E
a cube is divided into 125 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ?
"each side of the cube has 5 x 5 = 25 cubelets . only the interior cubelets are painted one colour . on each side , 3 x 3 = 9 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 9 = 54 the answer is d ."
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60
d
divide(subtract(125, multiply(multiply(const_4, const_2), const_3)), const_2)
multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)|
geometry
D
there are 190 items that are members of set u . of these items , 49 are members of set b , 59 are not members of either of set a or set b , and 23 are members of both sets a and b . how many of the members of set u are members of set a ?
"you had the answer almost right . the x = 82 refers to only set a . however what ' s being asked is how many members are part of set a . this will include : 1 . only set a 2 . set a and set b so the answer is set a = 82 + set ab = 82 + 23 = 105 d"
a ) 72 , b ) 85 , c ) 94 , d ) 105 , e ) 108
d
subtract(add(subtract(190, 59), 23), 49)
subtract(n0,n2)|add(n3,#0)|subtract(#1,n1)|
other
D
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 m 3 , then the rise in the water level in the tank will be :
"explanation : total volume of water displaced = ( 4 x 50 ) m 3 = 200 m 3 rise in water level = 200 / 40 Γ— 20 = 0.25 m = 25 cm answer : b"
a ) 20 cm , b ) 25 cm , c ) 35 cm , d ) 50 cm , e ) none of these
b
divide(multiply(4, 50), multiply(40, 20))
multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)|
physics
B
if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 22 , 28 , 19 , 33 and 35 ?
"it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 22 , 28 , 19 , 33 and 35 19 , 22 , 24 , 28 , 30 , 33 , 35 , 42 . . . numbers less than median are 19 , 22 , 24 , 28 . . the median of these numbers = center of 22 and 24 = 23 e"
a ) 33 , b ) 25 , c ) 27 , d ) 24 , e ) 23
e
divide(add(24, 28), const_2)
add(n1,n4)|divide(#0,const_2)|
general
E
suraj has a certain average of runs for 9 innings . in the 10 th innings he scores 200 runs thereby increasing his average by 8 runs . what is his average after the 10 th innings ?
"to improve his average by 8 runs per innings he has to contribute 9 x 8 = 72 runs for the previous 8 innings . thus , the average after the 9 th innings = 200 - 72 = 128 . answer : c"
a ) 149 , b ) 190 , c ) 128 , d ) 178 , e ) 190
c
divide(subtract(200, multiply(9, 8)), subtract(10, 9))
multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)|
general
C
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2340 , what will be b ' s share of the earnings ?
explanatory answer a , b and c will share the amount of $ 2340 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , the ratio of the work done by a : b : c when they work together will be equal to 1 / 6 : 1 / 8 : 1 / 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 / 6 : 24 / 8 : 24 / 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 2340 will be 4 : 3 : 2 . hence , b ' s share will be 3 * 2340 / 9 = 780 correct choice is ( c )
a ) $ 1100 , b ) $ 520 , c ) $ 780 , d ) $ 1170 , e ) $ 630
c
multiply(2340, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8)))))
inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)
physics
C
find 62976 Γ· ? = 123
"answer let 62976 / n = 123 then n = 62976 / 123 = 512 . option : c"
a ) 412 , b ) 502 , c ) 512 , d ) 522 , e ) none
c
divide(62976, 123)
divide(n0,n1)|
general
C
a tank is filled by 3 pipes with uniform flow . the first two pipes operating simultaneously fill the tan in the same time during which the tank is filled by the third pipe alone . the 2 nd pipe fills the tank 5 hours faster than first pipe and 4 hours slower than third pipe . the time required by first pipe is
if first pipe fills tank in x hrs , then second pipe fills tank in x - 5 hrs and 3 rd pipe fills tank in x - 9 hrs , then as per given condition 1 / x + 1 / ( x - 5 ) = 1 / ( x - 9 ) solving it , we get x = 15 hrs answer : c
a ) 13 hours , b ) 45 hours , c ) 15 hours , d ) 12 hours , e ) 11 hours
c
divide(add(multiply(add(5, 4), 2), sqrt(subtract(power(multiply(add(5, 4), 2), const_2), multiply(const_4, multiply(5, add(5, 4)))))), const_2)
add(n2,n3)|multiply(n1,#0)|multiply(n2,#0)|multiply(#2,const_4)|power(#1,const_2)|subtract(#4,#3)|sqrt(#5)|add(#1,#6)|divide(#7,const_2)
physics
C
john and david can finish a job together in 3 hours . if john can do the job by himself in 4 hours , what percent of the job does david do ?
"you can also plug in numbers . for example , bob and alice work at a donut factory and make 12 donuts which is the job ( i picked this as a smart number ) . john on his own works 12 / 4 = 3 donuts per hour . john and david work 12 / 3 = 4 donuts per hour so david works 1 donuts / hour to find out the percentage , david works 1 donuts / hr x 3 hours = 3 donuts per hour . therefore 3 donuts / 12 donuts = 1 / 4 = 25 % answer : e"
a ) 35 % , b ) 28 % , c ) 26 % , d ) 20 % , e ) 25 %
e
multiply(subtract(3, inverse(4)), const_100)
inverse(n1)|subtract(n0,#0)|multiply(#1,const_100)|
gain
E
rs . 200 amounts to rs . 800 in 8 years at simple interest . if the interest is increased by 5 % , it would amount to how much ?
"( 200 * 5 * 8 ) / 100 = 80 200 + 80 = 280 answer : d"
a ) 520 , b ) 440 , c ) 260 , d ) 280 , e ) 120
d
multiply(power(add(const_1, divide(5, const_100)), 8), 200)
divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|
gain
D
a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money t did bob receive ?
"notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ t = 20 answer : b"
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52
b
divide(multiply(32, const_2), const_3)
multiply(n2,const_2)|divide(#0,const_3)|
general
B
a train passes a platform in 22 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is
"speed of the train = 54 km / hr = ( 54 Γ— 10 ) / 36 m / s = 15 m / s length of the train = speed Γ— time taken to cross the man = 15 Γ— 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 12 = > 300 + l = 15 Γ— 22 = 330 = > l = 330 - 300 = 30 meter answer is d ."
a ) 40 , b ) 50 , c ) 60 , d ) 30 , e ) 20
d
multiply(multiply(const_0_2778, 54), subtract(22, 20))
multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)|
physics
D
by selling a book for 250 , 20 % profit was earned . what is the cost price of the book ?
"sp = 120 % of cp ; : . cp = 250 Γ— 100 / 120 = 208 option ' b '"
a ) a ) 215 , b ) b ) 208 , c ) c ) 230 , d ) d ) 235 , e ) e ) 240
b
original_price_before_gain(20, 250)
original_price_before_gain(n1,n0)|
gain
B
the speed at which a man can row a boat in still water is 6 km / hr . if he rows downstream , where the speed of current is 3 km / hr , how many seconds will he take to cover 80 meters ?
"the speed of the boat downstream = 6 + 3 = 9 km / hr 9 km / hr * 5 / 18 = 2.5 m / s the time taken to cover 80 meters = 80 / 2.5 = 32 seconds . the answer is b ."
a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44
b
divide(80, multiply(add(6, 3), const_0_2778))
add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
B
the diagonal of a rhombus are 40 m and 30 m . its area is :
area of the rhombus = 1 / 2 d 1 d 2 = ( 1 / 2 Γ£ β€” 40 Γ£ β€” 30 ) cm ( power ) 2 = 40 Γ£ β€” 15 = 600 cm ( power ) 2 answer is a .
['a ) 600', 'b ) 450', 'c ) 350', 'd ) 500', 'e ) 620']
a
rhombus_area(40, 30)
rhombus_area(n0,n1)
geometry
A
if p / q = 2 / 7 , then 2 p + q = ?
"let p = 2 , q = 7 then 2 * 2 + 7 = 11 so 2 p + q = 11 . answer : a"
a ) 11 , b ) 14 , c ) 13 , d ) 15 , e ) 16
a
add(multiply(2, 2), 7)
multiply(n0,n2)|add(n1,#0)|
general
A
if a is a positive integer , and if the units digit of a ^ 2 is 9 and the units digit of ( a + 1 ) ^ 2 is 4 , what is the units z digit of ( a + 2 ) ^ 2 ?
"i also got a . by punching in numers : z . . . 7 ^ 2 = . . . 9 . . . 8 ^ 2 = . . . 4 . . . 9 ^ 2 = . . . 1 . a"
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) c . 14
a
power(add(multiply(9, 2), 2), 2)
multiply(n0,n1)|add(n0,#0)|power(#1,n0)|
general
A
the positive integers m and n leave remainders of 2 and 3 , respectively , when divided by 6 . m > n . what is the remainder when m – n is divided by 6 ?
m is 2 more than a multiple of 6 ; n is 3 more than a multiple of 6 . m - n will be 1 less than a multiple of 6 ( 5 , 11 , 17 , . . . ) , therefore m - n will yield a remainder of 5 when divided by 6 . answer : e .
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
e
add(6, subtract(2, 3))
subtract(n0,n1)|add(n2,#0)
physics
E
the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 12 years and that of the other 9 students is 16 years . the age of the 15 th student is ?
"age of the 15 th student = [ 15 * 15 - ( 12 * 5 + 16 * 9 ) ] = ( 225 - 204 ) = 21 years . answer : c"
a ) 11 years , b ) 17 years , c ) 21 years , d ) 14 years , e ) 12 years
c
subtract(multiply(15, 15), add(multiply(5, 12), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
C
a shopkeeper buys two articles for rs . 1000 each and then sells them , making 50 % profit on the first article and 50 % loss on second article . find the net profit or loss percent ?
"profit on first article = 50 % of 1000 = 500 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : a"
a ) 500 , b ) 768 , c ) 276 , d ) 280 , e ) 279
a
multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 50)), 1000), multiply(divide(const_100, add(const_100, 50)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 50)), 1000), multiply(divide(const_100, add(const_100, 50)), 1000))), const_100)
add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#2)|divide(const_100,#0)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|multiply(#8,const_100)|divide(#9,#7)|multiply(#10,const_100)|
gain
A
if ( a + b ) = 4 , ( b + c ) = 5 and ( c + d ) = 3 , what is the value of ( a + d ) ?
"given a + b = 4 b + c = 5 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 4 - 5 + 3 = 2 . option d . . ."
a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 .
d
subtract(add(4, 3), 5)
add(n0,n2)|subtract(#0,n1)|
general
D
the sum of two numbers is 40 and their difference is 4 . the ratio of the numbers is
sol . let the numbers be x and y . then , x + y / x - y = 40 / 4 = 10 ⇔ ( x + y ) = 10 ( x - y ) ⇔ 9 x = 11 y ⇔ x / y = 11 / 9 . answer a
a ) 11 : 9 , b ) 11 : 18 , c ) 21 : 19 , d ) 21 : 19 , e ) none
a
divide(divide(subtract(40, subtract(divide(40, const_2), const_2)), const_2), divide(subtract(divide(40, const_2), const_2), const_2))
divide(n0,const_2)|subtract(#0,const_2)|divide(#1,const_2)|subtract(n0,#1)|divide(#3,const_2)|divide(#4,#2)
general
A
what is the 28 th digit to the right of the decimal point in the decimal form of 4 / 11 ?
"4 / 11 = 0.363636 . . . the even numbered positions in the decimal expansion are all 6 . the answer is d ."
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
d
divide(4, 11)
divide(n1,n2)|
general
D
8 , 24 , 12 , 36 , 18 , 54 , ( . . . . )
"8 Γ— 3 = 24 24 Γ· 2 = 12 12 Γ— 3 = 36 36 Γ· 2 = 18 18 Γ— 3 = 54 54 Γ· 2 = 27 answer is a"
a ) 27 , b ) 68 , c ) 107 , d ) 108 , e ) 28
a
subtract(negate(36), multiply(subtract(24, 12), divide(subtract(24, 12), subtract(8, 24))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
A
how many times will the digit 6 be written when listing the integers from 1 to 1000 ?
"many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 6 ) 3000 / 10 = 300 times . answer : e ."
a ) 150 , b ) 280 , c ) 310 , d ) 420 , e ) 300
e
multiply(multiply(multiply(1, const_10), const_10), const_3)
multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3)|
general
E
in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 76 per kg ?
"( 96 - 76 ) / ( 76 - 68 ) = 20 / 8 = 5 / 2 answer : e"
a ) 5 : 8 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 5 : 2
e
divide(divide(subtract(96, 76), subtract(96, 68)), subtract(const_1, divide(subtract(96, 76), subtract(96, 68))))
subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|
other
E
the simple interest on rs . 23 for 3 months at the rate of 5 paise per rupeeper month is
sol . s . i . = rs . [ 23 * 5 / 100 * 3 ] = rs . 3.45 answer a
a ) 3.45 , b ) 4.5 , c ) 2.25 , d ) 3.21 , e ) none
a
divide(multiply(multiply(23, 3), 5), const_100)
multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)
gain
A
car a runs at the speed of 50 km / hr and reaches its destination in 8 hours . car b runs at the speed of 65 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ?
"car a travels 50 Γ— 8 = 400 km car b travels 65 Γ— 4 = 260 km the ratio is 400 : 260 = 40 : 26 = 20 : 13 the answer is c ."
a ) 3 : 7 , b ) 4 : 9 , c ) 20 : 13 , d ) 5 : 7 , e ) 6 : 11
c
divide(multiply(50, 8), multiply(65, 4))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)|
physics
C
for any integer m greater than 1 , $ m denotes the product of all the integers from 1 to m , inclusive . how many prime numbers are there between $ 7 + 2 and $ 7 + 10 , inclusive ?
$ is basically a factorial of a number . so , we are asked to find the number of primes between 7 ! + 2 and 7 ! + 10 , inclusive . from each number 7 ! + k were 2 ≀ k ≀ 102 ≀ k ≀ 10 we can factor out k , thus there are no pries in the given range . for example : 7 ! + 2 = 2 ( 3 * 4 * 5 * 6 * 7 + 1 ) - - > a multiple of 2 , thus not a prime ; 7 ! + 3 = 3 ( 2 * 4 * 5 * 6 * 7 + 1 ) - - > a multiple of 3 , thus not a prime ; . . . 7 ! + 10 = 10 ( 3 * 4 * 6 * 7 + 1 ) - - > a multiple of 10 , thus not a prime . answer : a .
a ) none , b ) one , c ) two , d ) three , e ) four
a
add(subtract(add(add(add(1, 1), 2), 10), add(7, 7)), const_2)
add(n0,n0)|add(n2,n2)|add(n3,#0)|add(n5,#2)|subtract(#3,#1)|add(#4,const_2)
general
A
a batsman makes a score of 92 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning .
"let the average after 17 th inning = x . then , average after 16 th inning = ( x – 3 ) . ∴ 16 ( x – 3 ) + 92 = 17 x or x = ( 92 – 48 ) = 44 . answer d"
a ) 36 , b ) 39 , c ) 42 , d ) 44 , e ) none of the above
d
add(subtract(92, multiply(17, 3)), 3)
multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|
general
D
a train speeds past a pole in 10 seconds and a platform 100 m long in 30 seconds . its length is ?
"let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 100 / 30 = x / 10 x = 50 m . answer : d"
a ) 188 m , b ) 876 m , c ) 251 m , d ) 50 m , e ) 45 m
d
multiply(100, subtract(const_2, const_1))
subtract(const_2,const_1)|multiply(n1,#0)|
physics
D
find the area of circle whose radius is 7 m ?
the area of circle = pie * r ^ 2 = 22 / 7 * 7 * 7 = 154 sq m answer : e
['a ) 121 sq m', 'b ) 184 sq m', 'c ) 174 sq m', 'd ) 124 sq m', 'e ) 154 sq m']
e
circle_area(7)
circle_area(n0)
geometry
E
a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio t of revenue from car sales in 1996 to revenue from truck sales in 1996 ?
a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2
a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3
a
divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11)))
add(n3,const_100)|add(n6,const_100)|subtract(const_100,n1)|subtract(#0,#1)|subtract(#1,#2)|divide(#3,#4)|
other
A
a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 60 percent .
1 hr = 20 percent . thus 15 min = 5 percent . now to charge 60 percent 180 min . answer : d
a ) 145 , b ) 150 , c ) 175 , d ) 180 , e ) 130
d
multiply(divide(60, 20), const_60)
divide(n2,n1)|multiply(#0,const_60)|
physics
D
what is the value of ( p + q ) / ( p - q ) if p / q is 4 ?
"( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 4 + 1 ) / ( 4 - 1 ) = 5 / 3 = 5 / 3 answer : a"
a ) 5 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 7 / 8 , e ) 8 / 7
a
divide(multiply(add(add(const_100, const_60), const_1), 4), const_100)
add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)|
general
A
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 8 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 / 8 ) = 1 / 34 leak will empty the tank in 34 hrs . answer : d"
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 34 hr , e ) 36 hr
d
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 8), 1), 8))))
divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)|
physics
D
what percent is 50 gm of 1 kg ?
"1 kg = 1000 gm 50 / 1000 Γ— 100 = 5000 / 1000 = 5 % e )"
a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 5 %
e
multiply(divide(50, 1), const_100)
divide(n0,n1)|multiply(#0,const_100)|
gain
E
machine m , n , o working simultaneously machine m can produce x units in 3 / 4 of the time it takes machine n to produce the same amount of units . machine n can produce x units in 2 / 7 the time it takes machine o to produce that amount of units . if all 3 machines are working simultaneously , what fraction of the total output is produced by machine n ?
now ultimately the speed of every machine is given with respect to mach o . so lets assume the speed of o , say 12 hrs to make x units ( assuming 6 because we can see we will need to divide by 3 and 4 mach o makes x units in 12 hrs so , mach n = 2 / 7 of o = 2 / 7 * 12 = 24 / 7 hrs to make x units and mach m = 3 / 4 of n = 3 / 4 * 24 / 7 = 1 / 6 hrs to make x units no they are running simultaneously . lets see how much each mach makes in 1 hr mach o = x / 12 units mach n = 7 / 24 units mach m = x / 6 units in 1 hr , together they make - x / 12 + 7 / 24 + x / 6 = 13 / 24 so what ratio of this has mach n made ? ( 7 / 24 ) / ( 13 / 24 ) = 7 / 13 ans : b = 7 / 13
a ) 1 / 2 , b ) 7 / 13 , c ) 4 / 13 , d ) 8 / 29 , e ) 6 / 33
b
divide(7, subtract(multiply(7, 2), const_1))
multiply(n2,n3)|subtract(#0,const_1)|divide(n3,#1)
general
B
a regular hexagon is there . the mid points of the sides were joined and formed another hexagon . then what is the percentage reduction in area .
let abcdef be the regular hexagon with sides ab and ed parallel . p and q are mid points of af and ef resp . ae = root ( 3 ) * s ( s is side of hexagon ) then pq = ( root ( 3 ) * s ) / 2 mid point theorem now u know side of inner hexagon use formula area of hexagon = ( 3 * root ( 3 ) ) 2 * s ^ 2 and find diff in area of two hexagon multiply by 100 and divide the product by area of bigger hexagon ans 25 % answer : a
a ) 25 % , b ) 35 % , c ) 45 % , d ) 55 % , e ) 65 %
a
multiply(subtract(const_1, divide(const_3, const_4)), const_100)
divide(const_3,const_4)|subtract(const_1,#0)|multiply(#1,const_100)
general
A
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 2500 more than d , what is b ' s share ?
"let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 2500 = > x = 2500 . b ' s share = rs . 2 x = 2 * 2500 = rs . 5000 . answer : e"
a ) a ) 8239 , b ) b ) 2900 , c ) c ) 4500 , d ) d ) 2393 , e ) e ) 5000
e
multiply(multiply(subtract(4, 3), 2500), 3)
subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)|
general
E
two numbers are in the ratio of 1 : 2 . if 9 be added to both , their ratio changes to 3 : 5 . the greater number is
"let the ratio be x : y , given x / y = 1 / 2 , ( x + 9 ) / ( y + 9 ) = 3 / 5 = > x = 18 and y = 36 answer : e"
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36
e
multiply(1, 9)
multiply(n0,n2)|
other
E
pat , kate and mark charged a total of 180 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much times as mark , how many more hours did mark charge to the project than kate .
let kate charge for x hours , then pat charged for 2 x and mat - for 6 x . so , 2 x + 6 x + x = 180 - total hours charged for , x = 20 . mat charged 6 x - x or 5 x for more hours than kate , or for 100 hours . e is correct
a ) 18 , b ) 36 , c ) 72 , d ) 90 , e ) 100
e
multiply(divide(180, add(add(1, const_2), multiply(const_2, 3))), subtract(multiply(const_2, 3), 1))
add(n1,const_2)|multiply(n2,const_2)|add(#0,#1)|subtract(#1,n1)|divide(n0,#2)|multiply(#4,#3)
general
E
a student multiplied a number by 3 / 5 instead of 5 / 3 . what is the percentage error .
"explanation : let the number be x , then , 5 / 3 βˆ’ 3 / 5 = 16 / 15 x error % = ( 16 / 15 x βˆ— 3 / 5 βˆ— 100 ) option a"
a ) 64 % , b ) 65 % , c ) 66 % , d ) 67 % , e ) none of these
a
multiply(divide(subtract(divide(5, 3), divide(3, 5)), divide(5, 3)), const_100)
divide(n1,n0)|divide(n0,n1)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)|
general
A
if the cost price of 50 articles is equal to the selling price of 28 articles , then the gain or loss percent is ?
"given that , cost price of 50 article is equal to selling price of 28 articles . let cost price of one article = rs . 1 selling price of 28 articles = rs . 50 but cost price of 28 articles = rs . 28 therefore , the trader made profit . \ percentage of profit = 22 / 28 * 100 = 78.57 % answer : c"
a ) 22 , b ) 65 , c ) 78.57 , d ) 33 , e ) 25
c
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 28), 50)), divide(multiply(const_100, 28), 50)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
C
two right circular cylinders of equal volumes have their heights in the ratio 1 : 2 . find the ratio of their radii .
explanation : let their heights be h and 2 h and radii be r and r respectively then . Ο€ r 2 h = Ο€ r 2 ( 2 h ) = > r 2 / r 2 = 2 h / h = 2 / 1 = > r / r = √ 2 / 1 = > r : r = √ 2 : 1 answer is c
['a ) √ 3 : 1', 'b ) √ 7 : 1', 'c ) √ 2 : 1', 'd ) 2 : 1', 'e ) 3 : 1']
c
sqrt(2)
sqrt(n1)
geometry
C
the area of a parallelogram is 98 sq m and its altitude is twice the corresponding base . then the length of the base is ?
"2 x * x = 98 = > x = 7 answer : c"
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5
c
sqrt(divide(98, const_2))
divide(n0,const_2)|sqrt(#0)|
geometry
C
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ?
"x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,15 = 15 / 100 x = 80 * 15 / 100 x = 12 12 / 10 = 1,2 = 120 % , answer c"
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 %
c
multiply(divide(multiply(15, subtract(const_1, divide(20, const_100))), 10), const_100)
divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)|
gain
C
last year a certain bond price with a face value of 5000 yielded 8 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ?
"interest = 0.08 * 5000 = 0.065 * selling price - - > selling price = 0.08 * 5000 / 0.065 - - > selling price = ~ 6,154 answer : e ."
a ) 4063 , b ) 5325 , c ) 5351 , d ) 6000 , e ) 6154
e
divide(multiply(5000, divide(8, const_100)), divide(6.5, const_100))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)|
gain
E
fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 30 kg of fresh grapes ?
"the weight of non - water in 30 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 30 Γ’ Λ† β€” 0.1 = x Γ’ Λ† β€” 0.8 - - > x = 3.75 answer : a"
a ) 3.75 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these
a
multiply(divide(divide(multiply(subtract(const_100, 90), 30), const_100), subtract(const_100, 20)), const_100)
subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)|
gain
A
a team of 8 persons joins in a shooting competition . the best marksman scored 85 points . if he had scored 92 points , the average score for the team would have been 84 . the number of points , the team scored was
"let the total score be x . ( x + 92 - 85 ) / 8 = 84 . so , x + 7 = 672 = > x = 665 . answer : a"
a ) 665 , b ) 127 , c ) 272 , d ) 287 , e ) 227
a
subtract(add(multiply(84, 8), 85), 92)
multiply(n0,n3)|add(n1,#0)|subtract(#1,n2)|
general
A
the product of two numbers is 4107 . if the h . c . f of these numbers is 37 , then the greater number is :
"let the numbers be 37 a and 37 b . then , 37 a * 37 b = 4107 = > ab = 3 now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 * 1 , 37 * 3 ) i . e . , ( 1 , 111 ) . greater number = 111 . answer : c"
a ) 101 , b ) 107 , c ) 111 , d ) 112 , e ) 113
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
C
in town x , 64 percent of the population are employed , and 50 percent of the population are employed males . what percent of the employed people in town x are females ?
"we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 50 are employed males , hence 14 % are employed females . ( employed females ) / ( total employed people ) = 14 / 64 = 22 % answer : b ."
a ) 16 % , b ) 22 % , c ) 32 % , d ) 40 % , e ) 52 %
b
multiply(divide(subtract(64, 50), 64), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
B
in a shop , the profit is 320 % of the cost . if the cost increases by 25 % but the selling price remains constant , approximately what percentage of the selling price is the profit ?
"cp . = rs . 100 . = > profit = rs . 320 , s . p . = rs . 420 . new c . p . = 125 % of rs . 100 = rs . 125 new s . p . = rs . 420 . profit = rs . ( 420 - 125 ) = rs . 295 . req = ( 295 / 420 * 100 ) = = > 70 % answer a"
a ) 70 % , b ) 50 % , c ) 100 % , d ) 60 % , e ) 55 %
a
multiply(divide(subtract(add(320, const_100), add(25, const_100)), add(320, const_100)), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)|
gain
A
the average of squares of first 11 consecutive even numbers is :
2 ( n + 1 ) ( 2 n + 1 ) / 3 ( 2 * 12 * 23 ) / 3 = 2 * 4 * 23 = 184 answer : c
a ) 164 , b ) 174 , c ) 184 , d ) 194 , e ) 204
c
divide(multiply(multiply(const_2, add(11, const_1)), add(multiply(11, const_2), const_1)), const_3)
add(n0,const_1)|multiply(n0,const_2)|add(#1,const_1)|multiply(#0,const_2)|multiply(#2,#3)|divide(#4,const_3)
general
C
a pineapple costs rs 9 each and a watermelon costs rs . 6 each . if i spend rs 51 on total what is the number of pineapple i purchased ?
"5 * 9 + 6 = 51 5 pineapples answer : c"
a ) 6 , b ) 7 , c ) 5 , d ) 3 , e ) 2
c
divide(subtract(51, multiply(const_4, 9)), 6)
multiply(n0,const_4)|subtract(n2,#0)|divide(#1,n1)|
general
C
if the cost price of 10 articles is equal to the selling price of 20 articles , what is the % profit or loss made by the merchant ?
"let the cost price of 1 article be $ 1 . therefore , cost price of 10 articles = 10 * 1 = $ 10 the selling price of 20 articles = cost price of 10 articles = $ 10 . now , we know the selling price of 20 articles . let us find the cost price of 20 articles . cost price of 20 articles = 20 * 1 = $ 20 . therefore , profit made on sale of 20 articles = selling price of 20 articles - cost price of 20 articles = 10 - 20 = - $ 10 . as the profit is in the negative , the merchant has made a loss of $ 10 . therefore , % loss = loss / cp * 100 % loss = - 10 / 20 * 100 = 50 % loss . e"
a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 50 % loss
e
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 20), 10)), divide(multiply(const_100, 20), 10)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
E
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is :
"let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 9 = 4 : 3 . answer b"
a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these
b
divide(sqrt(16), sqrt(9))
sqrt(n1)|sqrt(n0)|divide(#0,#1)|
physics
B
a retailer sells 6 shirts . the first 2 he sells for $ 40 and $ 50 . if the retailer wishes to sell the 6 shirts for an overall average price of over $ 50 , what must be the minimum average price of the remaining 4 shirts ?
"first 2 shirts are sold for $ 40 and $ 50 = $ 90 . to get average price of $ 50 , total sale should be 6 * $ 50 = $ 300 so remaining 4 shirts to be sold for $ 300 - $ 90 = $ 210 answer should be 210 / 4 = $ 52.50 that is e"
a ) $ 37.00 , b ) $ 40.00 , c ) $ 45.50 , d ) $ 48.00 , e ) $ 52.50
e
divide(subtract(multiply(6, 50), add(40, 50)), subtract(6, 2))
add(n2,n3)|multiply(n0,n5)|subtract(n0,n1)|subtract(#1,#0)|divide(#3,#2)|
general
E
in the rectangular coordinate system , points ( 4 , 0 ) and ( – 4 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c
"it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 4 b"
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above
b
sqrt(power(4, const_2))
power(n0,const_2)|sqrt(#0)|
geometry
B
6 women can do 75 unit of work in 8 days by working 5 hr / day in how many days 4 women do 30 units of work by working 8 hr / day
( 6 x 8 x 5 ) / 75 = ( ax 4 x 8 ) / 30 a = 3 answer : a
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
divide(30, multiply(multiply(divide(divide(divide(75, 6), 8), 5), 8), 4))
divide(n1,n0)|divide(#0,n2)|divide(#1,n3)|multiply(n2,#2)|multiply(n4,#3)|divide(n5,#4)
physics
A
set x consists of 10 integers and has median of 30 and a range of 20 . what is the value of the greatest possible integer that can be present in the set ?
note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 20 , 50 this is possible . hence it is the greatest such number . answer ( e )
a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50
e
add(30, 20)
add(n1,n2)
general
E
if three painters can complete three rooms in two hours , how many hours 5 painters would it take to do 20 rooms ?
"explanation : three painters can complete three rooms in two hours . so 20 rooms can be painted in 8 hrs by 5 painters answer : e ) 5 painters"
a ) 6 , b ) 9 , c ) 7 , d ) 8 , e ) 5
e
divide(multiply(multiply(20, const_3), const_3), multiply(const_3, const_3))
multiply(n1,const_3)|multiply(const_3,const_3)|multiply(#0,const_3)|divide(#2,#1)|
physics
E
julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 120 and $ 126 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same , what was the amount of julie ' s initial savings ?
"$ 120 for 2 years = $ 60 per year . extra $ 6 yearned with the compound interest is the percent yearned on percent . so , $ 6 is yearned on $ 60 , which means that the interest = 10 % . this on the other hand means that half of the savings = 60 * 10 = $ 600 . twice of that = $ 1,200 . answer : d ."
a ) 600 , b ) 720 , c ) 1080 , d ) 1200 , e ) 1440
d
divide(120, divide(multiply(const_2, subtract(126, 120)), 120))
subtract(n1,n0)|multiply(#0,const_2)|divide(#1,n0)|divide(n0,#2)|
gain
D
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 440 sq . feet , how many feet of fencing will be required ?
"given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 440 breadth = 22 feet area to be fenced = 2 b + l = 2 ( 22 ) + 20 = 64 feet answer : c"
a ) 34 , b ) 40 , c ) 64 , d ) 88 , e ) 92
c
add(multiply(divide(440, 20), const_2), 20)
divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|
geometry
C