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a sum fetched a total simple interest of rs . 100 at the rate of 5 p . c . p . a . in 4 years . what is the sum ?
"sol . principal = rs . [ 100 * 100 / 5 * 4 ] = rs . [ 10000 / 20 ] = rs . 500 . answer c"
a ) 800 , b ) 600 , c ) 500 , d ) 1000 , e ) 300
c
divide(divide(multiply(100, const_100), 5), 4)
multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|
gain
an auction house charges a commission of 18 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ?
"say the price of the house was $ x , then 0.18 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 200,000 ( 18 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c ."
a ) $ 115,000 , b ) $ 160,000 , c ) $ 200,000 , d ) $ 240,000 , e ) $ 365,000
c
add(multiply(18, 10), 10)
multiply(n0,n2)|add(n2,#0)|
general
if one person completes a journey in 10 hrs . he travels 1 st half of the journey at the rate of 21 km / hr and 2 nd half at therate of 24 km / hr . find the total journey in km .
distance = speed x time let time taken to travel the first half = x hr then time taken to travel the second half = ( 10 - x ) hr distance covered in the the first half = 21 x distance covered in the the second half = 24 ( 10 - x ) but distance covered in the the first half = distance covered in the the second half = > ...
a ) 200 km , b ) 212 km , c ) 224 km , d ) 230 km , e ) 256 km
c
add(multiply(21, divide(10, 2)), multiply(24, divide(10, 2)))
divide(n0,n3)|multiply(n2,#0)|multiply(n4,#0)|add(#1,#2)
general
m = { - 6 , - 5 , - 4 , - 3 , - 2 , - 1 } t = { - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2 } if an integer is to be randomly selected from set m above and an integer is to be randomly selected from set t above , what is the probability that the product of the two integers will be negative ?
we will have a negative product only if 1 or 2 are selected from set t . p ( negative product ) = 2 / 8 = 1 / 4 the answer is b .
a ) 0 , b ) 1 / 4 , c ) 2 / 5 , d ) 1 / 2 , e ) 3 / 5
b
divide(2, multiply(4, 2))
multiply(n2,n4)|divide(n4,#0)
general
the radius of a cylinder is 6 m , height 21 m . the total surface area of the cylinder is ?
"total surface area of the cylinder is = 2 Ο€ r ( r + h ) = 2 Γ— 22 / 7 Γ— 6 ( 6 + 8 ) = 2 Γ— 22 / 7 Γ— 6 ( 14 ) = 44 Γ— 12 = 528 m ( power 2 ) answer is b ."
a ) 525 , b ) 528 , c ) 522 , d ) 529 , e ) 521
b
multiply(circumface(6), 21)
circumface(n0)|multiply(n1,#0)|
geometry
an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 2,160 , what was the total number of pairs of contact lenses that she sold ?
"( x + 5 ) * 150 + x * 85 = 2160 = > x = 6 total lens = 6 + ( 6 + 5 ) = 17 answer d"
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19
d
add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5)
add(n0,n1)|multiply(const_100,const_4)|multiply(n2,const_10)|multiply(n0,n2)|add(#1,#2)|add(n2,#4)|add(#5,const_1000)|subtract(#6,#3)|divide(#7,#0)|multiply(#8,const_2)|add(n2,#9)|
general
what is the tenth digit of ( 5 ! * 5 ! - 5 ! * 3 ! ) / 5 ?
( 5 ! * 5 ! + 5 ! * 3 ! ) / 5 = 5 ! ( 5 ! + 3 ! ) / 5 = 120 ( 120 + 6 ) / 5 = 3024 units digit of the above product will be equal to 2 answer b
a ) 1 , b ) 2 , c ) 0 , d ) 6 , e ) 7
b
floor(divide(divide(subtract(multiply(factorial(5), factorial(5)), multiply(factorial(5), factorial(3))), 5), multiply(const_100, const_10)))
factorial(n0)|factorial(n3)|multiply(const_10,const_100)|multiply(#0,#0)|multiply(#0,#1)|subtract(#3,#4)|divide(#5,n0)|divide(#6,#2)|floor(#7)
general
30 pens and 75 pencils were purchased for 570 . if the average price of a pencil was 2.00 , find the average price of a pen .
"since average price of a pencil = 2 ∴ price of 75 pencils = 150 ∴ price of 30 pens = ( 570 – 150 ) = 360 ∴ average price of a pen = 420 ⁄ 60 = 14 answer e"
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
e
divide(subtract(570, multiply(75, 2.00)), 30)
multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0)|
general
if 20 typists can type 42 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ?
"20 typists can type 42 letters , so 30 typists can type = 42 * 30 / 20 42 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 42 * 30 * 60 / 20 * 20 = 189 a is the answer"
a ) 189 , b ) 72 , c ) 144 , d ) 216 , e ) 400
a
multiply(divide(multiply(42, const_3), 20), 30)
multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)|
physics
a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates q apply to what fraction of a week ?
"number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates q / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c"
a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10
c
divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4)))
add(const_2,const_3)|add(const_3,const_4)|divide(n2,const_2)|multiply(n2,const_2)|multiply(#0,#2)|multiply(n2,#1)|add(#4,#3)|divide(#6,#5)|
physics
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 18 households had a both a car and a bike and 44 had a car , how many had bike only ?
"{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 18 + 11 - - > { bike } = 53 - - > # those who have bike only is { bike } - { both } = 53 - 18 = 35 . answer : b ."
a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10
b
subtract(subtract(add(subtract(90, 11), 18), 44), 18)
subtract(n0,n1)|add(n2,#0)|subtract(#1,n3)|subtract(#2,n2)|
other
two trains of equal are running on parallel lines in the same direction at 42 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ?
"let the length of each train be x m . then , distance covered = 2 x m . relative speed = 42 - 36 = 6 km / hr . = 6 * 5 / 18 = 5 / 3 m / sec . 2 x / 36 = 5 / 3 = > x = 30 . answer : a"
a ) 30 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m
a
divide(multiply(36, divide(multiply(subtract(42, 36), const_1000), const_3600)), const_2)
subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|
general
a customer pays 70 dollars for a coffee maker after a discount of 20 dollars what is the original price of the coffe maker ?
let x be the original price . x - 20 = 70 x - 20 + 20 = 70 + 20 x + 0 = 90 x = 90 answer is c
a ) 50 , b ) 40 , c ) 90 , d ) 60 , e ) 20
c
add(70, 20)
add(n0,n1)
gain
if 40 % of a certain number is 160 , then what is 50 % of that number ?
"explanation : 40 % = 40 * 4 = 160 50 % = 50 * 4 = 200 answer : option c"
a ) 270 , b ) 380 , c ) 200 , d ) 360 , e ) 290
c
multiply(divide(160, divide(40, const_100)), divide(50, const_100))
divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1)|
gain
a man can row his boat with the stream at 12 km / h and against the stream in 4 km / h . the man ' s rate is ?
"ds = 12 s = ? s = ( 12 - 4 ) / 2 = 4 kmph answer : e"
a ) 1 kmph , b ) 4 kmph , c ) 98 kmph , d ) 6 kmph , e ) 4 kmph
e
divide(subtract(12, 4), const_2)
subtract(n0,n1)|divide(#0,const_2)|
gain
the circumferences of two circles are 132 meters and 352 meters . find the difference between the areas of the larger and the smaller circles ?
"let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 132 and 2 ∏ l = 352 s = 132 / 2 ∏ and l = 352 / 2 ∏ difference between the areas = ∏ l ^ 2 - ∏ s ^ 2 = ∏ { 66 ^ 2 / ∏ ^ 2 - 132 ^ 2 / ∏ ^ 2 } = 66 ^ 2 / ∏ - 132 ^ 2 / ∏ = ( 66 - 132 ) ( 66 + 132 ) / ∏ = ( 110 ) ( 242 ) / ( 22 / 7...
a ) 2996 sq m , b ) 2897 sq m , c ) 4312 sq m , d ) 5768 sq m , e ) 8470 sq m
e
subtract(circle_area(divide(352, multiply(const_2, const_pi))), circle_area(divide(132, multiply(const_2, const_pi))))
multiply(const_2,const_pi)|divide(n1,#0)|divide(n0,#0)|circle_area(#1)|circle_area(#2)|subtract(#3,#4)|
geometry
the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if steve and dinley have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ?
"if steve and dinley submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combinedno discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 ...
a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54
b
divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2)
multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2)|
gain
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 16000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is :
"explanation : lot the total number of workers be v then , 8 ooov = ( 16000 * 7 ) + 6000 ( v - 7 ) < = > 2000 v = 70000 < = > v = 35 answer : c ) 35"
a ) 22 , b ) 21 , c ) 35 , d ) 37 , e ) 29
c
add(7, divide(multiply(7, subtract(16000, 8000)), subtract(8000, 6000)))
subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)|
general
when positive integer w is divided by 13 , the remainder is 2 . when n is divided by 8 , the remainder is 5 . how many such values are less than 180 ?
the equation that can be formed w is 13 x + 2 = 8 y + 5 . . 13 x - 3 = 8 y . . . as we can see x can take only odd values as the rhs will always be even . . also x can take values till 13 as 13 * 14 > 180 . . now we have to substitue x as 1 , 35 , 79 , 1113 . . . once we find 7 fitting in , any other value need not be ...
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
b
subtract(reminder(multiply(13, add(5, 2)), 8), 2)
add(n1,n3)|multiply(n0,#0)|reminder(#1,n2)|subtract(#2,n1)
general
a certain box has 11 cards and each card has one of the integers from 1 to 11 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on...
the average of the numbers is 6 the total number of ways to choose 2 cards from 11 cards is 11 c 2 = 55 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } , { 1,4 } , { 2,3 } the probability is 4 / 55 the answer is d .
a ) 1 / 11 , b ) 2 / 11 , c ) 2 / 33 , d ) 4 / 55 , e ) 5 / 66
d
divide(const_4, divide(factorial(11), multiply(factorial(2), factorial(subtract(11, 2)))))
factorial(n0)|factorial(n3)|subtract(n0,n3)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)|divide(const_4,#5)
general
if a 2 + b 2 + c 2 = 281 and ab + bc + ca = 4 , then a + b + c is
"by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 281 and ab + bc + ca = 4 , ( a + b + c ) ^ 2 = 281 + 2 ( 4 ) = 289 = 17 ^ 2 therefore : a + b + c = 17 answer : d"
a ) 16 , b ) 18 , c ) 22 , d ) 17 , e ) 20
d
sqrt(add(281, multiply(4, 2)))
multiply(n4,n0)|add(n3,#0)|sqrt(#1)|
general
pradeep has to obtain 25 % of the total marks to pass . he got 185 marks and failed by 25 marks . the maximum marks are
"explanation : let their maximum marks be x . then , 25 % of x = 185 + 25 = > 25 / 100 x = 210 x = ( 210100 / 25 ) x = 840 . answer : a"
a ) 840 , b ) 600 , c ) 800 , d ) 1000 , e ) 900
a
divide(add(185, 25), divide(25, const_100))
add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|
general
how many digits will be there to the right of the decimal point in the product of 98 and . 08216 ?
"product of 98 and . 08216 is 8.05168 . therefore number of digits to right of decimal point is 5 answer is a ."
a ) 5 , b ) 6 , c ) 9 , d ) 7 , e ) 8
a
subtract(subtract(const_100, 98), const_1)
subtract(const_100,n0)|subtract(#0,const_1)|
general
9 men went to a theater . 8 of them spent rs . 3 each over their tickets and the ninth spent rs . 2 more than the average expenditure of all the 9 . determine the total money spent by them ?
average of 9 = x 9 x = 8 * 3 + x * 2 x = 3.25 total = 9 * 3.25 = 29.25 b
a ) 29 , b ) 29.25 , c ) 31 , d ) 31.23 , e ) 32
b
multiply(divide(add(multiply(3, 8), 2), 8), 9)
multiply(n1,n2)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)
general
bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 48 white marbles . how many red marbles could be in bag a ?
"6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 48 = 6 k + 4 p = > k = 6 , p = 3 total red ' s in bag a will be 2 k = 12 e"
a ) 1 , b ) 3 , c ) 4 , d ) 10 , e ) 12
e
divide(48, add(multiply(3, 2), 4))
multiply(n1,n2)|add(n5,#0)|divide(n6,#1)|
other
a runs 1 2 / 3 times as fast as b . if a gives b a start of 80 m , how far must the winning post be so that a and b might reach it at the same time ?
explanation : ratio of the speeds of a and b = 5 : 1 = 5 : 3 thus , in race of 5 m , a gains 2 m over b . 2 m are gained by a in a race of 5 m . 80 m will be gained by a in race of 5 / 2 x 80 m = 200 m winning post is 200 m away from the starting point . answer is a
a ) 200 m , b ) 300 m , c ) 270 m , d ) 160 m , e ) 150 m
a
divide(multiply(add(multiply(1, 3), 2), 80), 2)
multiply(n0,n2)|add(n1,#0)|multiply(n3,#1)|divide(#2,n1)
general
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 18 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ?
"1 h - - - - - 3 ? - - - - - - 60 12 h rs = 18 + 21 = 39 t = 12 d = 39 * 12 = 468 answer : a"
a ) 468 , b ) 444 , c ) 676 , d ) 767 , e ) 663
a
add(multiply(divide(60, subtract(21, 18)), 18), multiply(divide(60, subtract(21, 18)), 21))
subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
physics
if d = 1 / ( 2 ^ 3 * 5 ^ 6 ) is expressed as a terminating decimal , how many nonzero digits will d have ?
"another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 6 ) = multiply and divide by 2 ^ 3 = 2 ^ 3 / ( 2 ^ 3 * 2 ^ 3 * 5 ^ 6 ) = 2 ^ 3 / 10 ^ 6 = > non zero digits are 8 = > ans a"
a ) one , b ) two , c ) three , d ) seven , e ) ten
a
add(1, 2)
add(n0,n1)|
general
if the perimeter and diagonal of a rectangle are 14 and 5 cms respectively , find its area .
in a rectangle , ( perimeter ) 2 / 4 = ( diagonal ) 2 + 2 Γ— area β‡’ ( 14 ) 2 / 4 = 5 ( 2 ) + 2 Γ— area 49 = 25 + 2 Γ— area ∴ area = 49 βˆ’ 25 / 2 = 24 / 2 = 12 cm 2 answer a
['a ) 12 cm 2', 'b ) 16 cm 2', 'c ) 20 cm 2', 'd ) 24 cm', 'e ) none of these']
a
divide(subtract(power(divide(14, const_2), const_2), power(5, const_2)), const_2)
divide(n0,const_2)|power(n1,const_2)|power(#0,const_2)|subtract(#2,#1)|divide(#3,const_2)
geometry
a train 110 m long is running with a speed of 50 km / h . in how many seconds will the train pass a man who is running at 5 km / h in the direction opposite to that in which the train is going ?
"the speed of the train relative to the man = 50 + 5 = 55 km / h . 55000 m / h * 1 h / 3600 s = ( 550 / 36 ) m / s ( 110 m ) / ( 550 / 36 m / s ) = ( 110 * 36 ) / 550 = 36 / 5 = 7.2 seconds the answer is b ."
a ) 6.1 , b ) 7.2 , c ) 8.3 , d ) 9.4 , e ) 10.5
b
divide(110, divide(add(50, 5), const_3_6))
add(n1,n2)|divide(#0,const_3_6)|divide(n0,#1)|
physics
when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in h hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of h ?
"rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * h = 1 = > x = 1 / h when they both work together then their rates get added up combined rate = ( w + x ) r * t = ...
a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2
b
add(subtract(4, 2), divide(const_1, add(2, 3)))
add(n0,n1)|subtract(n2,n0)|divide(const_1,#0)|add(#2,#1)|
physics
if two - third of a bucket is filled in 100 seconds then the time taken to fill the bucket completely will be .
2 / 3 filled in 100 seconds 1 / 3 filled in 50 secs then 2 / 3 + 1 / 3 = 100 + 50 seconds = 150 seconds answer : b
a ) 90 seconds , b ) 150 seconds , c ) 60 seconds , d ) 100 seconds , e ) 120 seconds
b
multiply(divide(100, const_2), const_3)
divide(n0,const_2)|multiply(#0,const_3)
physics
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 digit of ( a + 2 ) ^ 2 ?
"for unit digit of a ^ 2 to be 9 . . . unit digit of a has to be 3 or 7 . . . now for unit digit of ( a + 1 ) ^ 2 to be 4 . . unit digit of a has to be 1 or 7 . . . . from the above two conditions , unit value of a has to be 7 , which will satisfy both the conditions . . . now id unit digit of a is 7 , unit digit of ( ...
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 14
a
power(add(multiply(9, 2), 2), 2)
multiply(n0,n1)|add(n0,#0)|power(#1,n0)|
general
if the cost price is 81 % of selling price then what is the profit percentage .
selling price = rs 100 : then cost price = rs 81 : profit = rs 19 . profit = { ( 19 / 81 ) * 100 } % = 23.45 % answer is b .
a ) 22.45 , b ) 23.45 , c ) 32.45 , d ) 23.54 , e ) 23.55
b
multiply(divide(subtract(const_100, 81), 81), const_100)
subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)
gain
farm tax is levied on the 60 % of the cultivated land . the tax department collected total $ 5000 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :
"this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 5000 ) x 100 = 9.6 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 9.6 % x ( 100 / 60 ) = 16 % the corre...
a ) 15 % , b ) 16 % , c ) 0.125 % , d ) 0.2083 % , e ) none
b
divide(multiply(multiply(divide(480, 5000), const_100), const_100), 60)
divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|
general
in company j , the total monthly payroll for the 15 factory workers is $ 30000 and the total monthly payroll for the 30 office workers is $ 75000 . by how much does the average ( arithmetic mean ) monthly salary of an office worker exceed that of a factory worker in this company ?
the average monthly salary of a factory worker is : $ 30000 / 15 = $ 2000 . the average monthly salary of an office worker is : $ 75000 / 30 = $ 2500 . the difference in average salary is : $ 2500 - $ 2000 = $ 500 . the answer is b .
a ) $ 450 , b ) $ 500 , c ) $ 600 , d ) $ 650 , e ) $ 750
b
divide(divide(30000, 30), const_2)
divide(n1,n2)|divide(#0,const_2)
general
the average mark of the students of a class in a particular exam is 70 . if 5 students whose average mark in that exam is 50 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam ?
"let the number of students who wrote the exam be x . total marks of students = 70 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 70 x - ( 5 * 50 ) = 90 ( x - 5 ) 200 = 20 x = > x = 10 answer : c"
a ) 20 , b ) 15 , c ) 10 , d ) 12 , e ) 25
c
divide(subtract(multiply(90, 5), multiply(5, 50)), subtract(90, 70))
multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|
general
what is the smallest positive integer nn such that √ 6,480 βˆ— n is a perfect cube ?
"sol : let ' s factorize 6480 and we get 6480 = 3 ^ 4 * 2 ^ 4 * 5 now we need to see for what minimum value of n √ n * 6480 = a ^ 3 where a is an integer so from 6480 we already have 2 ^ 4 * 3 ^ 4 * 5 * n √ n = ( 2 ^ 2 ) ^ 3 * ( 3 ^ 2 ) ^ 3 * ( 5 ) ^ 3 why cause a is an integer which will need to be have the same facto...
a ) 5 , b ) 5 ^ 2 , c ) 30 , d ) 30 ^ 2 , e ) 30 ^ 4
e
add(const_3, const_4)
add(const_3,const_4)|
geometry
what will be the fraction of 12 %
"explanation : 12 * 1 / 100 = 3 / 25 . option d"
a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 3 / 25 , e ) none of these
d
divide(circle_area(divide(12, const_2)), const_2)
divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)|
gain
at joel ’ s bookstore , the current inventory is 40 % historical fiction . of the historical fiction books , 40 % are new releases , while 20 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ?
"let there be 100 books in all historic fiction books = 40 % of total = 40 other books = 60 new historic fiction = 40 % of 40 = 16 other new books = 20 % of 60 = 12 total new books = 28 fraction = 16 / 28 = 8 / 14 ans : d"
a ) 4 / 25 , b ) 8 / 23 , c ) 2 / 5 , d ) 8 / 14 , e ) 2 / 3
d
divide(divide(multiply(40, 40), const_100), add(divide(multiply(40, 40), const_100), divide(multiply(20, subtract(const_100, 40)), const_100)))
multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)|
gain
a person purchases 90 clocks and sells 40 clocks at a gain of 10 % and 50 clocks at a gain of 20 % . if he sold all of them at a uniform profit of 15 % , then he would have got rs . 40 less . the cost price of each clock is ?
"let c . p . of clock be rs . x . then , c . p . of 90 clocks = rs . 90 x . [ ( 110 % of 40 x ) + ( 120 % of 50 x ) ] - ( 115 % of 90 x ) = 40 44 x + 60 x - 103.5 x = 40 0.5 x = 40 = > x = 80 answer : c"
a ) 26 , b ) 28 , c ) 80 , d ) 26 , e ) 21
c
divide(40, subtract(add(multiply(40, add(const_1, divide(10, const_100))), multiply(50, add(const_1, divide(20, const_100)))), multiply(90, add(const_1, divide(15, const_100)))))
divide(n2,const_100)|divide(n4,const_100)|divide(n5,const_100)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|multiply(n1,#3)|multiply(n3,#4)|multiply(n0,#5)|add(#6,#7)|subtract(#9,#8)|divide(n1,#10)|
gain
a rectangle having length 120 cm and width 50 cm . if the length of the rectangle is increased by ten percent then how much percent the breadth should be decreased so as to maintain the same area .
"explanation : solution : ( 10 / ( 120 + 10 ) * 100 ) % = 7.69 % answer : d"
a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 7.69 % , e ) none of these
d
multiply(add(const_1, divide(divide(multiply(120, 50), add(120, divide(multiply(multiply(const_3, const_10), 120), const_100))), 50)), const_10)
multiply(n0,n1)|multiply(const_10,const_3)|multiply(n0,#1)|divide(#2,const_100)|add(n0,#3)|divide(#0,#4)|divide(#5,n1)|add(#6,const_1)|multiply(#7,const_10)|
geometry
the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 39 degrees . find the temperature on friday ?
"m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 39 tu + w + th = 192 - 39 = 153 f = 184 – 153 = 31 answer : c"
a ) 65 degrees , b ) 73 degrees , c ) 31 degrees , d ) 34 degrees , e ) 74 degrees
c
subtract(39, subtract(multiply(48, const_4), multiply(46, const_4)))
multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)|
general
it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 3 mile / hour , what is the speed with which joey returns ?
"let his speed for one half of the journey be 2 miles an hour let the other half be x miles an hour now , avg speed = 3 mile an hour 2 * 2 * x / 2 + x = 3 4 x = 3 x + 6 = > x = 6 b"
a ) 1 , b ) 6 , c ) 13 , d ) 14 , e ) 15
b
divide(2, subtract(divide(multiply(const_2, 2), 3), 1))
multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)|
physics
a man is 16 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is
"explanation : let ' s son age is x , then father age is x + 16 . = > 2 ( x + 2 ) = ( x + 16 + 2 ) = > 2 x + 4 = x + 18 = > x = 14 years option b"
a ) 21 years , b ) 14 years , c ) 16 years , d ) 18 years , e ) 26 years
b
divide(subtract(16, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))
multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|
general
a man ' s speed with the current is 15 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is ?
"man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 15 - 2.5 = 12.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km / hr answer is a ."
a ) 10 , b ) 20 , c ) 50 , d ) 30 , e ) 40
a
subtract(subtract(15, 2.5), 2.5)
subtract(n0,n1)|subtract(#0,n1)|
gain
a man ' s speed with the current is 21 km / hr and the speed of the current is 4.3 km / hr . the man ' s speed against the current is
"man ' s rate in still water = ( 21 - 4.3 ) km / hr = 16.7 km / hr . man ' s rate against the current = ( 16.7 - 4.3 ) km / hr = 12.4 km / hr . answer : d"
a ) 9.2 , b ) 10.3 , c ) 11.5 , d ) 12.4 , e ) 13
d
subtract(subtract(21, 4.3), 4.3)
subtract(n0,n1)|subtract(#0,n1)|
gain
three cubes of metal whose edges are 9 , 12 and 15 cm respectively , are melted and one new cube is made . find the edge of the new cube ?
"93 + 123 + 153 = a 3 = > a = 18 answer : d"
a ) 21 cm , b ) 19 cm , c ) 32 cm , d ) 18 cm , e ) 28 cm
d
power(add(add(volume_cube(9), volume_cube(12)), volume_cube(15)), const_0_33)
volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)|
physics
martin buys a pencil and a notebook for 80 cents . at the same store , gloria buys a notebook and an eraser for 85 cents , and zachary buys a pencil and an eraser for 45 cents . how many cents would it cost to buy 3 pencils , 3 notebooks , and 3 erasers ? ( assume that there is no volume discount . )
pencil + notebook = 80 notebook + eraser = 85 pencil + eraser = 45 let ' s add all three equations . 2 pencils + 2 notebooks + 2 erasers = 210 cents the cost to buy 3 of each would be ( 3 / 2 ) ( 210 ) = 315 the answer is b .
a ) 300 , b ) 315 , c ) 330 , d ) 345 , e ) 360
b
add(add(multiply(subtract(45, subtract(80, divide(add(85, subtract(80, 45)), const_2))), 3), multiply(subtract(80, divide(add(85, subtract(80, 45)), const_2)), 3)), multiply(divide(add(85, subtract(80, 45)), const_2), 3))
subtract(n0,n2)|add(n1,#0)|divide(#1,const_2)|multiply(n3,#2)|subtract(n0,#2)|multiply(n3,#4)|subtract(n2,#4)|multiply(n3,#6)|add(#7,#5)|add(#8,#3)
gain
a man buys a cycle for rs . 1400 and sells it at a loss of 25 % . what is the selling price of the cycle ?
"since , c . p = 1400 loss % = ( c . p - s . p ) / c . p * 100 25 = ( 1400 - s . p ) / 1400 * 100 so , after solving answer = 1050 . answer : a"
a ) s . 1050 , b ) s . 1160 , c ) s . 1190 , d ) s . 1202 , e ) s . 1204
a
divide(multiply(subtract(const_100, 25), 1400), const_100)
subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|
gain
1 + 3 = 2 2 + 3 = 10 3 + 3 = 30 4 + 3 = 68 5 + 3 = ? ?
1 ^ 3 + 3 - 2 = 2 , 2 ^ 3 + 3 - 1 = 10 , 3 ^ 3 + 3 - 0 = 30 , 4 ^ 3 + 3 + 1 = 68 , 5 ^ 3 + 3 + 2 = 130 answer : c
a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150
c
multiply(add(multiply(5, 5), 1), 5)
multiply(n12,n12)|add(n0,#0)|multiply(n12,#1)
general
a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 90 square feet of the wall , and no tiles ov...
"the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 90 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ...
a ) 160 , b ) 240 , c ) 360 , d ) 495 , e ) 560
d
add(90, multiply(divide(multiply(90, 3), const_2), 3))
multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)|
geometry
what is the difference between the place value of 2 in the numeral 7669 ?
answer : option ' e ' 600 - 60 = 540
a ) 160 , b ) 165 , c ) 180 , d ) 190 , e ) 540
e
subtract(multiply(multiply(2, const_3), const_100), multiply(multiply(2, const_3), const_10))
multiply(n0,const_3)|multiply(#0,const_100)|multiply(#0,const_10)|subtract(#1,#2)
general
machine a and machine b are each used to manufacture 770 sprockets . it takes machine a 10 hours longer to produce 770 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ?
"machine b : takes x hours to produce 770 sprockets machine a : takes ( x + 10 ) hours to produce 770 sprockets machine b : in 1 hour , b makes 770 / x sprockets machine a : in 1 hour , a makes 770 / ( x + 10 ) sprockets equating : 1.1 ( 770 / ( x + 10 ) ) = 770 / x 847 / ( x + 10 ) = 770 / x 847 x = 770 x + 7700 77 x ...
a ) 5 , b ) 7 , c ) 9 , d ) 10 , e ) 12
b
divide(770, divide(multiply(multiply(10, 770), divide(add(const_100, 10), const_100)), subtract(multiply(770, divide(add(const_100, 10), const_100)), 770)))
add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)|
gain
if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 370 will be :
z share = z , y = 1.2 z x = 1.25 Γ£ β€” 1.2 z , x + y + z = 740 ( 1.25 Γ£ β€” 1.2 + 1.2 + 1 ) z = 37 3.7 z = 370 , z = 100 answer : . c
a ) rs . 300 , b ) rs . 200 , c ) rs . 100 , d ) rs . 350 , e ) none of these
c
divide(370, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1))
divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,#2)|add(#2,#4)|add(#5,const_1)|divide(n2,#6)
general
village x has a population of 70000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ?
"let the population of two villages be equal after p years then , 70000 - 1200 p = 42000 + 800 p 2000 p = 28000 p = 14 answer is a ."
a ) 14 , b ) 19 , c ) 11 , d ) 18 , e ) 13
a
divide(subtract(70000, 42000), add(800, 1200))
add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|
general
a , b , c can do a piece of work in 11 days , 20 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by b and c on alternate days ?
a + b 1 day work = 1 / 11 + 1 / 20 = 31 / 220 a + c 1 day work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 days = 31 / 220 + 6 / 55 = 55 / 220 = 1 / 4 1 / 4 work is done by a in 2 days whole work will be done in 2 * 4 = 8 days answer is c
a ) 2 , b ) 4 , c ) 8 , d ) 10 , e ) 12
c
divide(const_2, add(add(divide(const_1, 11), divide(const_1, 20)), add(divide(const_1, 11), divide(const_1, 55))))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(const_2,#5)
physics
a group of men decided to do a work in 8 days , but 3 of them became absent . if the rest of the group did the work in 10 days , find the original number of men ?
"original number of men = 3 * 10 / ( 10 - 8 ) = 15 answer is a"
a ) 15 , b ) 20 , c ) 30 , d ) 25 , e ) 18
a
divide(multiply(3, 10), subtract(10, 8))
multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|
physics
x , y , and z are different prime numbers . the product x ^ 2 * y ^ 2 * z ^ 2 is divisible by how many different positive numbers ?
"the exponents of x ^ 2 * y ^ 2 * z ^ 2 are 2 , 2 , and 2 . the number of factors is ( 2 + 1 ) ( 2 + 1 ) ( 2 + 1 ) = 27 the answer is c ."
a ) 9 , b ) 18 , c ) 27 , d ) 36 , e ) 45
c
subtract(power(2, const_4), const_4)
power(n0,const_4)|subtract(#0,const_4)|
general
a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 36 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ?
"yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 36 hours . if the company used t...
a ) 3 , b ) 4 , c ) 6 , d ) 9 , e ) 12
d
divide(const_1, multiply(36, add(divide(const_1, 36), divide(const_1, const_2.0))))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(const_2.0,#2)|divide(const_1,#3)|
gain
? x 24 = 173 x 240
"let y x 24 = 173 x 240 then y = ( 173 x 240 ) / 24 = 173 x 10 = 1730 answer : d"
a ) 545 , b ) 685 , c ) 865 , d ) 1730 , e ) 534
d
divide(multiply(173, 240), 24)
multiply(n1,n2)|divide(#0,n0)|
general
in how many no . between 10 and 20 exactly two of the digits is 1 ?
"it ' s simple can be solved by elimination of answer choices . option a and b are too large , not possible . even ce are large to have correct choice . ans : d"
a ) 25 , b ) 35 , c ) 10 , d ) 1 , e ) 15
d
divide(divide(20, 10), 1)
divide(n1,n0)|divide(#0,n2)|
general
a man sells a car to his friend at 11 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was :
explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β€” 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 11 % . now s . p = rs . 45000 and loss = 11 % c . p . rs . [ 100 / 89 Γ£ β€” 45000 ] = rs . 50561.80 . correct option : c
a ) rs . 25561.80 , b ) rs . 37500.80 , c ) rs . 50561.80 , d ) rs . 60000 , e ) none of these
c
divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 11))
add(n2,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|divide(#1,#0)|multiply(#3,const_100)|divide(#4,#2)
gain
given that 268 x 74 = 19532 , find the value of 2.68 x . 74 .
"solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 Γ— . 74 = 1.9532 answer a"
a ) 1.9532 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none
a
multiply(divide(268, const_100), divide(74, const_100))
divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|
general
if a = { 1 , 3 , 5 } , b = { 3 , 5 , 6 } . find a βˆͺ b
"a = { 1 , 3,5 } b = { 3 , 5,6 } therefore , correct answer : a βˆͺ b = { 1 , 3 , 5 , 6 } b"
a ) { 1,5 } , b ) { 1,3 , 5,6 } , c ) { 2,6 } , d ) { 8,9 } , e ) { 4,12 }
b
add(divide(add(add(add(add(1, 3), 5), 3), 5), add(const_4, const_1)), 5)
add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|add(n2,#5)|
general
the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 3 . the sum of numbers is :
"let the numbers be 1 x and 3 x . then , their l . c . m = 3 x . so , 3 x = 48 or x = 16 . the numbers are 16 and 48 . hence , required sum = ( 16 + 48 ) = 64 . answer : e"
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 64
e
divide(multiply(1, 48), 3)
multiply(n0,n1)|divide(#0,n2)|
other
after a storm deposits 115 billion gallons of water into the city reservoir , the reservoir is 80 % full . if the original contents of the reservoir totaled 245 billion gallons , the reservoir was approximately what percentage full before the storm ?
"when the storm deposited 115 billion gallons , volume of water in the reservoir = 245 + 115 = 360 billion gallons if this is only 80 % of the capacity of the reservoir , the total capacity of the reservoir = 360 / 0.8 = 450 billion gallons therefore percentage of reservoir that was full before the storm = ( 245 / 450 ...
a ) 45 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 %
c
multiply(divide(245, divide(add(115, 245), divide(80, const_100))), const_100)
add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|divide(n2,#2)|multiply(#3,const_100)|
general
the sum of the ages of 5 children born at the intervals of 2 years each is 50 years . what is the age of the youngest child ?
let x = the youngest child . each of the other four children will then be x + 2 , x + 4 , x + 6 , x + 8 . we know that the sum of their ages is 50 . so , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 50 therefore the youngest child is 6 years old answer : a
a ) 6 , b ) 18 , c ) 10 , d ) 99 , e ) 38
a
divide(subtract(divide(50, divide(5, const_2)), multiply(subtract(5, const_1), 2)), const_2)
divide(n0,const_2)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,const_2)
general
an inspector rejects 0.08 % of the meters as defective , how many meters he examine to reject 2 meteres
explanation : it means that 0.08 % of x = 2 = > ( 8 / 100 Γ— 100 Γ— x ) = 2 = > x = 2 Γ— 100 Γ— 100 / 8 = > x = 2500 option d
a ) 1200 , b ) 2400 , c ) 1400 , d ) 2500 , e ) none of these
d
divide(multiply(2, const_100), 0.08)
multiply(n1,const_100)|divide(#0,n0)
gain
jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 36 / 49 , what...
"the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 36 / 49 = ( 6 / 7 ) * ( 6 / 7 ) therefore probability that the next one she draws is pink = 1 / 7 option ( a )"
a ) 1 / 7 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49
a
subtract(const_1, sqrt(divide(36, 49)))
divide(n0,n1)|sqrt(#0)|subtract(const_1,#1)|
general
increasing the original price of a certain item by 30 percent and then increasing the new price by 30 percent is equivalent to increasing the original price by what percent ?
"we ' re told that the original price of an item is increased by 30 % and then that price is increased by 30 % . . . . if . . . . starting value = $ 100 + 30 % = 100 + . 30 ( 100 ) = 130 + 30 % = 130 + . 30 ( 130 ) = 130 + 39 = 169 the question asks how the final price relates to the original price . this is essentiall...
a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 69.0
e
multiply(subtract(multiply(add(divide(30, const_100), const_1), add(divide(30, const_100), const_1)), const_1), const_100)
divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
the area of a square garden is a square feet and the perimeter is p feet . if a = 2 p + 15 , what is the perimeter of the garden , in feet ?
"perimeter of square = p side of square = p / 4 area of square = ( p ^ 2 ) / 16 = a given that a = 2 p + 15 ( p ^ 2 ) / 16 = 2 p + 15 p ^ 2 = 32 p + 240 p ^ 2 - 32 p - 240 = 0 p ^ 2 - 40 p + 6 p - 240 = 0 p ( p - 40 ) + 6 ( p + 40 ) = 0 ( p - 40 ) ( p + 6 ) = 0 p = 40 or - 6 discarding negative value , p = 40 answer is...
a ) 28 , b ) 36 , c ) 40 , d ) 56 , e ) 64
c
subtract(subtract(add(const_10, multiply(15, 2)), const_0_25), const_0_25)
multiply(n0,n1)|add(#0,const_10)|subtract(#1,const_0_25)|subtract(#2,const_0_25)|
geometry
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 370 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ?
explanation : let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 370 ) = 2 + 2 ( extra ) = > 2 x - 740 = x = > x = 740 . answer : d
a ) 237 , b ) 287 , c ) 197 , d ) 740 , e ) 720
d
multiply(370, const_2)
multiply(n0,const_2)
general
if the perimeter of a rectangular garden is 900 m , its length when its breadth is 190 m is ?
"2 ( l + 190 ) = 900 = > l = 260 m answer : d"
a ) 338 m , b ) 778 m , c ) 200 m , d ) 260 m , e ) 971 m
d
subtract(divide(900, const_2), 190)
divide(n0,const_2)|subtract(#0,n1)|
physics
two pipes can fill a tank in 20 minutes and 15 minutes . an outlet pipe can empty the tank in 10 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ?
"let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 20 + v / 15 - v / 10 = v / 60 per minute the tank will be filled in 60 minutes . the answer is e ."
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60
e
subtract(add(divide(const_1, 20), divide(const_1, 15)), divide(const_1, 10))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|
physics
brenda and sally run in opposite direction on a circular track , starting at diametrically opposite points . they first meet after brenda has run 100 meters . they next meet after sally has run 200 meters past their first meeting point . each girl runs at a constant speed . what is the length of the track in meters ?
"nice problem . + 1 . first timetogetherthey run half of the circumference . second timetogetherthey run full circumference . first time brenda runs 100 meters , thus second time she runs 2 * 100 = 200 meters . since second time ( when they run full circumference ) brenda runs 200 meters and sally runs 200 meters , thu...
a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 500
d
add(multiply(const_2, 100), 200)
multiply(n0,const_2)|add(n1,#0)|
physics
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 11 ) , what is the value of x ?
"( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 11 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 11 hence x = 13 . answer is c"
a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17
c
add(11, 2)
add(n0,n5)|
general
the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 2 ) and ( 9 , 2 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ?
if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 1 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 1 ) is 1 / 5 . the answer is d .
a ) 5 , b ) 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 8
d
divide(const_1, divide(add(subtract(9, 1), const_2), const_2))
subtract(n2,n0)|add(#0,const_2)|divide(#1,const_2)|divide(const_1,#2)
general
for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - third of his earnings to charity . if the worker ' s charitable contributions totaled one - eigh...
"let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 3 = 1 / 8 ( 11 x + nx ) ; 11 / 10 + n / 3 = 1 / 8 ( 11 + n ) 33 + 10 n / 30 = 11 + n / 8 = > 264 + 80 n = 330 + 30 n = > 50 n = 66 n = 66 / 50 = 33 / 25 answer : d ."
a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 33 / 25 , e ) 4
d
divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3))))
add(const_1,const_4)|inverse(const_10)|subtract(const_12,const_1)|subtract(const_12,const_3)|inverse(#3)|inverse(#0)|subtract(#4,#1)|subtract(#5,#4)|multiply(#2,#6)|divide(#8,#7)|
general
what is the units digit of ( 5 ! * 4 ! + 6 ! * 5 ! ) / 3 ?
"( 5 ! * 4 ! + 6 ! * 5 ! ) / 3 = 5 ! ( 4 ! + 6 ! ) / 3 = 120 ( 24 + 720 ) / 3 = ( 120 * 744 ) / 3 = 120 * 248 units digit of the above product will be equal to 0 answer d"
a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 1
d
divide(add(multiply(factorial(5), factorial(4)), multiply(factorial(5), factorial(5))), 5)
factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)|
general
two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 1 , respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio 4 : 3 . what is the ratio of the composition of the two basic elements in alloy x ...
mixture a has a total of 5 + 3 = 8 parts . if in the final mixture this represents 4 parts , then the total number of parts in mixture b should be ( 8 / 4 ) * 3 = 6 . so , we should take of mixture b a quantity with 3 and 3 parts , respectively . this will give us in the final mixture ( 5 + 3 ) : ( 3 + 3 ) , which mean...
a ) 1 : 1 , b ) 2 : 3 , c ) 5 : 2 , d ) 4 : 3 , e ) 7 : 9
d
divide(add(multiply(4, 5), multiply(3, divide(add(5, 3), const_2))), add(multiply(4, 3), multiply(3, divide(add(5, 3), const_2))))
add(n0,n1)|multiply(n0,n4)|multiply(n1,n4)|divide(#0,const_2)|multiply(n1,#3)|add(#1,#4)|add(#2,#4)|divide(#5,#6)
other
if x is 11 percent greater than 70 , then x =
"11 % of 70 = ( 70 * 0.11 ) = 7.7 11 % greater than 70 = 70 + 7.7 = 77.7 answer is clearly a ."
a ) 77.7 , b ) 91.0 , c ) 88.0 , d ) 70.9 , e ) 71.2
a
add(70, multiply(divide(11, const_100), 70))
divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|
general
if an examination 63 % of the candidates in english , 65 % passed in mathematics , and 27 % failed in both subjects . what is the pass percentage ?
fail in english = 100 - 63 = 37 % fail in maths = 100 - 65 = 35 % so pass % = 100 - ( 37 + 35 - 27 ) = 55 % answer : a
a ) 55 % , b ) 60 % , c ) 65 % , d ) 75 % , e ) none .
a
subtract(const_100, subtract(add(subtract(const_100, 63), subtract(const_100, 65)), 27))
subtract(const_100,n0)|subtract(const_100,n1)|add(#0,#1)|subtract(#2,n2)|subtract(const_100,#3)
gain
a train running at the speed of 60 km / hr crosses a pole in 15 seconds . find the length of the train ?
"speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 15 = 250 meter answer : a"
a ) 250 meter , b ) 876 meter , c ) 167 meter , d ) 719 meter , e ) 169 meter
a
multiply(divide(multiply(60, const_1000), const_3600), 15)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
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 1 strokes ?
"left after 1 st stroke = 1 / 2 so removed = 1 - 1 / 2 = 1 / 2"
a ) 1 / 2 , 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
fred and sam are standing 40 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 4 miles per hour and sam walks at a constant speed of 4 miles per hour , how many miles has sam walked when they meet ?
"relative distance = 40 miles relative speed = 4 + 4 = 8 miles per hour time taken = 40 / 8 = 5 hours distance travelled by sam = 4 * 5 = 20 miles = c"
a ) 5 , b ) 9 , c ) 20 , d ) 30 , e ) 45
c
multiply(4, divide(40, add(4, 4)))
add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|
physics
the average of first seven multiples of 5 is :
"explanation : ( 5 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 5 x 28 / 7 = 20 answer : a"
a ) 20 , b ) 16 , c ) 15 , d ) 8 , e ) 10
a
add(5, const_1)
add(n0,const_1)|
general
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot at the rate of 26.50 per metre is 5,300 , what is the length of the plot ( in metres ) ?
"perimeter of the rectangular plot = [ ( b + 20 ) + b ] Γ— 2 = 5300 / 26.5 = 200 = 200 ∴ ( 2 b + 20 ) 2 = 200 β‡’ b = 40 β‡’ l = 40 + 20 = 60 m answer e"
a ) 40 , b ) 120 , c ) 50 , d ) data inadequate , e ) none of these
e
subtract(divide(divide(5,300, 26.50), const_2), multiply(const_2, 20))
divide(n2,n1)|multiply(n0,const_2)|divide(#0,const_2)|subtract(#2,#1)|
gain
a , b and c are entered into a partnership . a invested rs . 6500 for 6 months , b invested rs . 8400 for 5 months and c invested for rs . 10000 for 3 months . a is a working partner and gets 5 % of the total profit for the same . find the share of c in a total profit of rs . 7400 ?
65 * 6 : 84 * 5 : 100 * 3 26 : 28 : 20 c share = 74000 * 95 / 100 = 7030 * 20 / 74 = > 1900 answer : b
a ) 2998 , b ) 1900 , c ) 2788 , d ) 2662 , e ) 1122
b
divide(multiply(subtract(7400, divide(multiply(5, 7400), const_100)), multiply(10000, 3)), add(add(multiply(6500, 6), multiply(8400, 5)), multiply(10000, 3)))
multiply(n4,n5)|multiply(n3,n7)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#1,const_100)|add(#4,#0)|subtract(n7,#5)|multiply(#0,#7)|divide(#8,#6)
gain
if 25 men do a work in 96 days , in how many days will 40 men do it ?
"25 * 96 = 40 * x x = 60 days answer : d"
a ) 66 , b ) 53 , c ) 55 , d ) 60 , e ) 61
d
divide(multiply(25, 96), 40)
multiply(n0,n1)|divide(#0,n2)|
physics
there are 8 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed .
"first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 8 and for second = 7 so total methods = 8 * 7 = 56 answer : d"
a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57
d
add(choose(8, 2), choose(8, 2))
choose(n0,n1)|add(#0,#0)|
probability
the expression ( 12.86 Γ— 12.86 + 12.86 Γ— p + 0.14 Γ— 0.14 ) will be a perfect square for p equal to
explanation : 12.86 Γ— 12.86 + 12.86 Γ— p + 0.14 Γ— 0.14 = ( 12.86 ) 2 + 12.86 Γ— p + ( 0.14 ) 2 this can be written as ( 12.86 + 0.14 ) 2 = 132 , if 12.86 Γ— p = 2 Γ— 12.86 Γ— 0.14 i . e . , if p = 2 Γ— 0.14 = 0.28 hence , p = 0.28 . answer : option a
a ) 0.28 , b ) 0.26 , c ) 1 , d ) 0 , e ) 2
a
multiply(0.14, const_2)
multiply(n3,const_2)
general
if the sample interest on a sum of money 20 % per annum for 2 years is $ 400 , find the compound interest on the same sum for the same period at the same rate ?
"rate = 20 % time = 2 years s . i . = $ 400 principal = 100 * 400 / 20 * 2 = $ 1000 amount = 1000 ( 1 + 20 / 100 ) ^ 2 = $ 1440 c . i . = 1440 - 1000 = $ 440 answer is c"
a ) $ 460 , b ) $ 510 , c ) $ 440 , d ) $ 500 , e ) $ 550
c
subtract(add(divide(multiply(add(divide(multiply(400, const_100), multiply(20, 2)), divide(multiply(divide(multiply(400, const_100), multiply(20, 2)), 20), const_100)), 20), const_100), add(divide(multiply(400, const_100), multiply(20, 2)), divide(multiply(divide(multiply(400, const_100), multiply(20, 2)), 20), const_1...
multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_100)|add(#2,#4)|multiply(n0,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|
gain
a pupil ' s marks were wrongly entered as 83 instead of 63 . due to that the average marks for the class got increased by half . the number of pupils in the class is
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 . x / 2 = ( 83 - 63 ) = > x / 2 = 20 = > x = 40 . answer : b"
a ) 36 , b ) 40 , c ) 99 , d ) 13 , e ) 12
b
multiply(subtract(83, 63), const_2)
subtract(n0,n1)|multiply(#0,const_2)|
general
a man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km per hour . find the distance travelled on foot .
total time = 7 hrs let the distance travelled by foot @ 8 kmph be x kms ? distance travlled by bicycle @ 16 kmph be ( 80 - x ) kms atq . 7 hr = x / 8 + ( 80 - x ) / 16 ? 7 = ( 2 x + 80 - x ) / 16 ? x = 32 kms answer : b .
a ) 26 km , b ) 32 km , c ) 30 km , d ) 28 km , e ) none
b
subtract(multiply(16, 7), 80)
multiply(n1,n3)|subtract(#0,n0)
physics
after decreasing 24 % in the price of an article costs rs . 820 . find the actual cost of an article ?
"cp * ( 76 / 100 ) = 820 cp = 10.78 * 100 = > cp = 1079 answer : b"
a ) 1400 , b ) 1079 , c ) 1200 , d ) 1023 , e ) 1523
b
divide(820, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
if x ^ 2 – x = 2 , then one possible value of x – 4 =
x ^ 2 – x = 2 i . e . x ^ 2 – x - 2 = 0 i . e . x ^ 2 + x - 2 x - 2 = 0 i . e . ( x - 2 ) ( x + 1 ) = 0 i . e . x = 2 or - 1 i . e . x - 4 = 2 - 4 or - 1 - 4 i . e . x - 4 = - 2 or - 5 answer : option b
a ) - 9 , b ) - 5 , c ) - 3 , d ) - 1 , e ) 5
b
subtract(divide(add(const_1, sqrt(add(const_1, multiply(const_4, 2)))), const_2), 4)
multiply(n0,const_4)|add(#0,const_1)|sqrt(#1)|add(#2,const_1)|divide(#3,const_2)|subtract(#4,n2)
general
a volunteer organization is recruiting new members . in the fall they manage to increase their number by 8 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ?
( 100 % + 8 % ) * ( 100 % - 19 % ) = 1.08 * . 81 = 0.8748 1 - 0.8748 = 12.52 % lost = - 12.52 % the answer is e the organization has lost 12.52 % of its total volunteers from fall to spring .
a ) 16.16 % , b ) 15.15 % , c ) 14.14 % , d ) 13.33 % , e ) 12.52 %
e
subtract(const_100, multiply(multiply(add(const_1, divide(8, const_100)), subtract(const_1, divide(19, const_100))), const_100))
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)
general
the side of a cube is 12 m , find the lateral surface area ?
lateral surface = 4 a ( power ) 2 4 Γ— 12 ( power ) 2 = 4 Γ— 144 = > 516 m ( power ) 2 answer is b .
['a ) 816', 'b ) 516', 'c ) 716', 'd ) 216', 'e ) 916']
b
rectangle_area(add(multiply(const_4, const_10), const_3), 12)
multiply(const_10,const_4)|add(#0,const_3)|rectangle_area(n0,#1)
geometry
a train passes a platform in 25 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 = 20 = > 300 + l = 15 Γ— 25 = 375 = > l = 375 - 300 = 75 meter answer...
a ) 75 , b ) 25 , c ) 26 , d ) 23 , e ) 22
a
multiply(multiply(const_0_2778, 54), subtract(25, 20))
multiply(n2,const_0_2778)|subtract(n0,n1)|multiply(#0,#1)
physics