Problem
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| options
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if the average of 6 digits is 16 and the average of 4 of them is 10 , calculate the average of the remaining 2 numbers ?
|
"explanation : dividend on 1 share = ( 20.5 * 50 ) / 100 = rs . 10.25 rs . 25 is income on an investment of rs . 100 rs . 10.25 is income on an investment of rs . ( 10.25 * 100 ) / 25 = rs . 41 answer : e"
|
a ) 25 , b ) 66 , c ) 18 , d ) 19 , e ) 41
|
e
|
divide(multiply(divide(multiply(20.5, 50), const_100), const_100), 25)
|
multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)|
|
gain
|
a company pays 20.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ?
|
a 59 - sided convex polygon has 59 vertices . if we examine a single vertex , we can see that we can connect it with 56 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 59 vertices can be connected with 56 vertices to create a diagonal then the total number of diagonals would be ( 59 ) ( 56 ) = 3304 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3304 by 2 to get 1652 . the answer is b .
|
a ) 1168 , b ) 1652 , c ) 2452 , d ) 3304 , e ) 4256
|
b
|
divide(factorial(59), multiply(factorial(subtract(59, const_2)), factorial(const_2)))
|
factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)
|
general
|
how many diagonals does a 59 - sided convex polygon have ?
|
"let c . p . be $ 100 . then , s . p . = $ 133 let marked price be $ x . then , 95 / 100 x = 133 x = 13300 / 95 = $ 140 now , s . p . = $ 140 , c . p . = $ 100 profit % = 40 % . a"
|
a ) 140 , b ) 120 , c ) 130 , d ) 110 , e ) 150
|
a
|
multiply(const_100, divide(add(const_100, 33), subtract(const_100, 5)))
|
add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|
|
gain
|
a shopkeeper sold an book offering a discount of 5 % and earned a profit of 33 % . what would have been the percentage of profit earned if no discount was offered ?
|
"swim in still water at = 5 speed of river = 3 us = 5 - 3 = 2 distance = 10 t = 10 / 2 = 5 answer : c"
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
|
c
|
divide(10, subtract(5, 3))
|
subtract(n2,n1)|divide(n0,#0)|
|
gain
|
calculate how long it will take a swimmer to swim a distance of 10 km against the current of a river which flows at 3 km / hr , given that he can swim in still water at 5 km / h
|
"cost price = 900 profit = 30 % = 30 % of 900 = 270 selling price = cp + profit sp = 1170 a discount of 30 % to employees means 30 % off on 1170 so 30 % of 1170 = 351 ans a"
|
a ) 352 , b ) 356 , c ) 358 , d ) 362 , e ) 365
|
a
|
divide(add(divide(multiply(900, 30), const_100), 900), multiply(divide(900, const_100), const_2))
|
divide(n1,const_100)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,const_2)|add(n1,#2)|divide(#4,#3)|
|
gain
|
a computer store offers employees a 30 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 900 dollars and marked up the price 30 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 30 % off retail price ) as opposed to the final retail price .
|
"let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 6360 [ 4 l - 40 ] = 6360 / 26.50 = 240 4 l = 280 l = 280 / 4 = 70 meters . answer : a"
|
a ) 70 , b ) 200 , c ) 300 , d ) 400 , e ) 140
|
a
|
subtract(divide(divide(6360, 26.50), const_2), multiply(const_2, 20))
|
divide(n2,n1)|multiply(n0,const_2)|divide(#0,const_2)|subtract(#2,#1)|
|
physics
|
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 6360 , what is the length of the plot in metres ?
|
"the # of factors 240 has is 20 ; so out of 20 factors only four are multiples of 15 : 15 , 30 , 60 , 120 and 240 , itself ; so , the probability is 5 / 20 = 1 / 4 . answer : e ."
|
a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4
|
e
|
divide(divide(choose(15, const_1), 15), power(const_3, const_2))
|
choose(n1,const_1)|power(const_3,const_2)|divide(#0,n1)|divide(#2,#1)|
|
other
|
each factor of 240 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 15 ?
|
"let us first write the numbers in the form of prime factors : 15 = 3 * 5 22 = 2 * 11 24 = 2 * 17 ^ 1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 1320 option d"
|
a ) 60 , b ) 120 , c ) 240 , d ) 1320 , e ) 1720
|
d
|
lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 22)
|
add(const_10,const_2)|multiply(const_10,const_3)|subtract(#1,const_3)|lcm(#0,#2)|lcm(n1,#3)|
|
general
|
what is the least common multiple of 15 , 22 and 24 ?
|
explanation : let the total score be x . ( x + 92 - 85 ) / 8 = 84 . so , x + 7 = 672 = > x = 665 . answer : a ) 665
|
a ) 665 , b ) 376 , c ) 998 , d ) 1277 , e ) 1991
|
a
|
subtract(add(multiply(84, 8), 85), 92)
|
multiply(n0,n3)|add(n1,#0)|subtract(#1,n2)
|
general
|
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 :
|
"one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 3 , then 3 x = y , where x is the smaller integer . if adding 10 to the smaller integer makes the ratio 2 to 1 , then x + 10 = 2 y . substituting y = 3 x into the second equation yields x + 10 = 6 x . so , x = 2 ( smaller integer ) and , y = 3 x = 6 ( larger integer ) so a is the correct answer . another option is to test the answer choices . a ) the larger number is 6 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 2 if 10 is added to the smaller number , the ratio becomes 2 to 1 if we add 10 to the smaller number ( 2 ) , we get 12 . so , the new ratio is 12 to 6 which is the same as 2 to 1 . therefore choice a is correct . b ) the larger number is 9 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 3 if 10 is added to the smaller number , the ratio becomes 2 to 1 if we add 10 to the smaller number ( 3 ) , we get 13 so , the new ratio is 13 to 9 . no good . we want a resulting ratio of 2 to 1 eliminate b c ) the larger number is 10 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 3.33 ( which is not an integer ) since both numbers are integers , the larger number can not be 10 . eliminate c d ) the larger number is 12 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 4 if 10 is added to the smaller number , the ratio becomes 2 to 1 if we add 10 to the smaller number ( 4 ) , we get 14 so , the new ratio is 14 to 12 which is 7 to 6 . no good . we want a resulting ratio of 2 to 1 eliminate d e ) the larger number is 15 given : the two integers are in the ratio of 1 to 3 so , the smaller number must be 5 if 10 is added to the smaller number , the ratio becomes 2 to 1 if we add 10 to the smaller number ( 5 ) , we get 15 so , the new ratio is 15 to 15 . this is the same as the ratio of 1 to 1 eliminate e therefore a is the only correct choice ."
|
a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15
|
a
|
multiply(divide(multiply(1, 10), subtract(3, 1)), 3)
|
multiply(n2,n4)|subtract(n1,n4)|divide(#0,#1)|multiply(n1,#2)|
|
other
|
two integers are in the ratio of 1 to 3 . if 10 is added to the smaller number , the ratio becomes 2 to 1 . find the larger integer .
|
"explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 20 / 5 = 4 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 × 4 = 44 secs . answer : c"
|
a ) 47 , b ) 76 , c ) 44 , d ) 66 , e ) 11
|
c
|
multiply(divide(20, subtract(6, const_1)), subtract(12, const_1))
|
subtract(n0,const_1)|subtract(n3,const_1)|divide(n2,#0)|multiply(#2,#1)|
|
physics
|
at 6 ′ o a clock ticks 6 times . the time between first and last ticks is 20 seconds . how long does it tick at 12 ′ o clock
|
"money paid in cash = rs . 1000 balance payment = ( 30000 - 1000 ) = rs . 29000 answer : c"
|
a ) 22678 , b ) 26699 , c ) 29000 , d ) 19000 , e ) 26711
|
c
|
subtract(30000, 1000)
|
subtract(n0,n2)|
|
gain
|
the price of a t . v . set worth rs . 30000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ?
|
"sol . let the length of the wire b h . radius = 1 / 2 mm = 1 / 20 cm . then , 22 / 7 * 1 / 20 * 1 / 20 * h = 55 ⇔ = [ 55 * 20 * 20 * 7 / 22 ] = 7000 cm = 70 m . answer d"
|
a ) 84 m , b ) 88 m , c ) 120 m , d ) 70 m , e ) none
|
d
|
divide(55, multiply(power(divide(1, const_2), const_2), const_pi))
|
divide(n1,const_2)|power(#0,const_2)|multiply(#1,const_pi)|divide(n0,#2)|
|
physics
|
55 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length of the wire in metres will be :
|
"l = ( 48 - 16 ) m = 32 m , b = ( 34 - 16 ) m = 18 m , h = 8 m . volume of the box = ( 32 x 18 x 8 ) m 3 = 4608 m 3 . answer : option c"
|
a ) 4830 , b ) 5120 , c ) 4608 , d ) 7500 , e ) 8960
|
c
|
volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(34, multiply(8, const_2)), 8)
|
multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|
|
geometry
|
a metallic sheet is of rectangular shape with dimensions 48 m x 34 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is :
|
"let the length of the train be x meters . the speed of the train is x / 15 . then , x + 120 = 25 * ( x / 15 ) 10 x = 1800 x = 180 meters the answer is d ."
|
a ) 100 m , b ) 140 m , c ) 130 m , d ) 180 m , e ) 170 m
|
d
|
multiply(120, subtract(const_2, const_1))
|
subtract(const_2,const_1)|multiply(n1,#0)|
|
physics
|
a train speeds past a pole in 15 seconds and a platform 120 meters long in 25 seconds . what is the length of the train ( in meters ) ?
|
some multiple of 7 + some multiple of 5 should yield 63 . to get to a some multiple of 5 , we should ensure that a 3 or 8 ( 5 + 3 ) should be a multiple of 7 . 63 is a direct multiple of 7 , however in this case there wo n ' t be any guava . hence the next option is to look for a multiple of 7 that has 8 as the unit digit . 28 satisfies this hence no . of apples is 4 and no of bananas is 7 . c
|
a ) 12 , b ) 13 , c ) 11 , d ) 14 , e ) 5
|
c
|
add(divide(subtract(6.3, multiply(0.7, const_4)), 0.5), const_4)
|
multiply(n0,const_4)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_4)
|
other
|
a certain fruit stand sold apples for $ 0.70 each and guava for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ?
|
"we know 25 % people study biology , therefore the no of people not studying = 100 - 25 = 75 % > therefore the people not studying biology out of a total 880 people are = 75 % of 880 > ( 75 / 100 ) * 880 = 660 people e"
|
a ) 500 , b ) 600 , c ) 620 , d ) 640 , e ) 660
|
e
|
multiply(divide(880, const_100), subtract(const_100, 25))
|
divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|
|
other
|
if 25 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ?
|
"let the smaller number be x . then larger number = ( x + 1380 ) . x + 1380 = 6 x + 15 5 x = 1365 x = 273 large number = 273 + 1380 = 1653 e"
|
a ) 1245 , b ) 1345 , c ) 1455 , d ) 1577 , e ) 1653
|
e
|
multiply(divide(subtract(1380, 15), subtract(6, const_1)), 6)
|
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|
|
general
|
find large number from below question the difference of two numbers is 1380 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder
|
"the total number of elementary events associated to the random experiments of throwing four dice simultaneously is : = 6 × 6 × 6 × 6 = 64 = 6 × 6 × 6 × 6 = 64 n ( s ) = 64 n ( s ) = 64 let xx be the event that all dice show the same face . x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 , 5,5 ) , ( 6,6 , 6,6 ) } x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 , 5,5 ) , ( 6,6 , 6,6 ) } n ( x ) = 6 n ( x ) = 6 hence required probability , = n ( x ) n ( s ) = 664 = n ( x ) n ( s ) = 664 = 1 / 216 c"
|
a ) 2 / 113 , b ) 3 / 117 , c ) 1 / 216 , d ) 3 / 111 , e ) 4 / 121
|
c
|
multiply(divide(const_3, add(const_3, const_3)), divide(const_3, add(const_3, const_3)))
|
add(const_3,const_3)|divide(const_3,#0)|multiply(#1,#1)|
|
probability
|
4 dice are thrown simultaneously on the board . find the probability which show the same face ?
|
official solution : ( b ) we know that a # b = 100 and a # b = 4 a ² + 4 b ² + 8 ab . so 4 a ² + 4 b ² + 8 ab = 100 we can see that 4 a ² + 4 b ² + 8 ab is a well - known formula for ( 2 a + 2 b ) ² . therefore ( 2 a + 2 b ) ² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbers . so we can conclude that 2 ( a + b ) = 10 . ( a + b ) + 8 = 10 / 2 + 8 = 13 . the correct answer is d
|
a ) 5 , b ) 8 , c ) 10 , d ) 13 , e ) 17
|
d
|
add(sqrt(divide(100, 4)), 8)
|
divide(n6,n0)|sqrt(#0)|add(n4,#1)
|
general
|
tough and tricky questions : word problems . operation # is defined as : a # b = 4 a ^ 2 + 4 b ^ 2 + 8 ab for all non - negative integers . what is the value of ( a + b ) + 8 , when a # b = 100 ?
|
"n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 88 , there are 44 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 44 + 12 ) / 88 = 66 / 88 = 33 / 44 ans is e"
|
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 33 / 44
|
e
|
divide(add(divide(88, 2), divide(88, 8)), 88)
|
divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|
|
general
|
if an integer n is to be chosen at random from the integers 1 to 88 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?
|
"its e . total juice rquired = 240 * 6 = 1440 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1440 / 48 = 30 . answer e"
|
a ) a ) 25 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 30
|
e
|
divide(divide(multiply(240, 6), 12), const_4)
|
multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)|
|
general
|
according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 240 6 ounces servings of orange juice ?
|
"∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 38400 ⁄ 12 = 3200 answer c"
|
a ) 4038 , b ) 8076 , c ) 3200 , d ) 4845.6 , e ) none of these
|
c
|
multiply(divide(divide(38400, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100))
|
multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)|
|
general
|
mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 38400 , how much amount per month would he have spent on clothes and transport ?
|
total cost of the items he purchased = rs . 50 given that out of this rs . 50 , 90 paise is given as tax = > total tax incurred = 90 paise = rs . 90 / 100 let the cost of the tax free items = x given that tax rate = 15 % ∴ ( 50 − 90 / 100 − x ) 15 / 100 = 90 / 100 ⇒ 15 ( 50 − 0.9 − x ) = 90 ⇒ ( 50 − 0.9 − x ) = 6 ⇒ x = 50 − 0.9 − 6 = 43.1 c
|
a ) a ) 19.7 , b ) b ) 20 , c ) c ) 43.1 , d ) d ) 21.5 , e ) e ) 22
|
c
|
subtract(subtract(50, divide(90, const_100)), divide(90, 15))
|
divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)
|
gain
|
daniel went to a shop and bought things worth rs . 50 , out of which 90 paise went on sales tax on taxable purchases . if the tax rate was 15 % , then what was the cost of the tax free items ?
|
"answer given expression = ( 7 + 7 + 7 ÷ 7 ) / ( 5 + 5 + 5 ÷ 5 ) = ( 14 + 1 ) / ( 10 + 1 ) = 15 / 11 option : c"
|
a ) 1 , b ) 1 / 5 , c ) 15 / 11 , d ) 3 / 11 , e ) none
|
c
|
subtract(divide(multiply(7, add(7, const_1)), const_2), divide(multiply(subtract(7, const_1), 7), const_2))
|
add(n3,const_1)|subtract(n0,const_1)|multiply(n3,#0)|multiply(n0,#1)|divide(#2,const_2)|divide(#3,const_2)|subtract(#4,#5)|
|
general
|
( 7 + 7 + 7 ÷ 7 ) / ( 5 + 5 + 5 ÷ 5 ) = ?
|
"explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 3410 * ( 6 / 31 ) = 110 * 6 = 660 . answer : option a"
|
a ) 660 , b ) 700 , c ) 800 , d ) 900 , e ) none of these
|
a
|
multiply(3410, divide(6, add(add(add(10, add(4, const_1)), 10), 6)))
|
add(n0,const_1)|add(n2,#0)|add(n2,#1)|add(n1,#2)|divide(n1,#3)|multiply(n3,#4)|
|
general
|
if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 3410 , r will receive
|
let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 1 * 6 = 6 days to produce 1 w widgets . answer : b
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
|
b
|
divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2)
|
multiply(n2,n3)|multiply(n0,#0)|subtract(#1,#0)|divide(#2,n0)
|
general
|
running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 1 w widgets ?
|
"let number of each type of coin = x . then , 1 × x + . 50 × x + . 25 x = 175 ⇒ 1.75 x = 175 ⇒ x = 100 coins answer d"
|
a ) 20 coins , b ) 30 coins , c ) 50 coins , d ) 100 coins , e ) none of these
|
d
|
divide(175, add(add(inverse(const_4), inverse(const_2)), const_1))
|
inverse(const_4)|inverse(const_2)|add(#0,#1)|add(#2,const_1)|divide(n2,#3)|
|
general
|
a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 175 , how many coins of each type are there ?
|
solution suppose y invested rs . y then , 5000 / y = 2 / 6 â € ¹ = â € º y = ( 5000 ã — 6 / 2 ) . â € ¹ = â € º y = 15000 . answer d
|
a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 15,000 , e ) none
|
d
|
divide(multiply(5000, 6), 2)
|
multiply(n1,n2)|divide(#0,n0)
|
gain
|
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 6 . if x invested rs . 5000 . the amount invested by y is
|
"this implies 9 a - b = 10 b + 70 , 9 a - b = - 12 b - 2 a , 10 b + 70 = - 12 b - 2 a manipulating the second equation gives us 9 a - b = 10 b + 70 = = > 9 a - 11 b = 70 answer is b"
|
a ) - 4 , b ) 70 , c ) 0 , d ) 2 , e ) 4
|
b
|
multiply(negate(multiply(divide(70, 2), 2)), 9)
|
divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)|
|
general
|
if 9 a - b = 10 b + 70 = - 12 b - 2 a , what is the value of 9 a - 11 b ?
|
answer : option e
|
a ) 100000 , b ) 222222 , c ) 333333 , d ) 888888 , e ) 999999
|
e
|
square_area(const_pi)
|
square_area(const_pi)|
|
general
|
the largest number of six digits
|
"s = 600 / 40 = 15 mps s = 3600 / 15 = 240 sec = 4 min . answer : a"
|
a ) 4 min , b ) 2 min , c ) 8 min , d ) 9 min , e ) 3 min
|
a
|
divide(divide(multiply(3, const_1000), speed(600, 40)), const_60)
|
multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)|
|
physics
|
a 600 meter long train crosses a signal post in 40 seconds . how long will it take to cross a 3 kilometer long bridge , at the same speed ?
|
if x is a positive integer , the lowest value would be x = 1 , because 0 is not included in the natural numbers . that ' s a technicality the gmat would not expect students to know . if x = 1 , then the expression equals - 14 . answer = ( b ) .
|
a ) - 15 , b ) - 14 , c ) - 13 , d ) - 12 , e ) - 11
|
b
|
subtract(subtract(const_1, const_1), 15)
|
subtract(const_1,const_1)|subtract(#0,n1)
|
general
|
what is the least possible value of x ^ 2 - 15 where x is a positive number .
|
john spent and gave to his two friends a total of 3.25 + 2.20 + 2.20 = $ 7.65 money left 16.10 - 7.65 = $ 8.45 answer : e
|
a ) $ 6.45 , b ) $ 8.35 , c ) $ 8.75 , d ) $ 8.85 , e ) $ 8.45
|
e
|
subtract(16.1, add(3.25, add(2.2, 2.2)))
|
add(n2,n2)|add(n1,#0)|subtract(n0,#1)
|
general
|
little john had $ 16.10 . he spent $ 3.25 on sweets and gave to his two friends $ 2.20 each . how much money was left ?
|
"3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 10 pumps then 48 / 10 = 4.8 hrs . answer : a"
|
a ) 4.8 , b ) 3.6 , c ) 1.1 , d ) 1.2 , e ) 1.3
|
a
|
divide(multiply(multiply(3, 8), 2), 10)
|
multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|
|
physics
|
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 10 pumps work to empty the tank in 1 day ?
|
given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 / pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 / pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 / 2 = 6 ( curved part , not including the flat rounded base ) . but the total surface area = 6 + area of the base of a hemisphere . = 6 + 3 = 9 . answer is d ! !
|
['a ) 6 / π', 'b ) 9 / π', 'c ) 6', 'd ) 9', 'e ) 12']
|
d
|
add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi)))
|
divide(n2,const_pi)|multiply(n0,const_pi)|multiply(#0,#1)|multiply(#0,const_pi)|divide(#2,n1)|add(#4,#3)
|
geometry
|
the surface area of a sphere is 4 π r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area e of that hemisphere ?
|
"5354 x 51 = 5354 x ( 50 + 1 ) = 5354 x 50 + 5354 x 1 = 267700 + 5354 = 273054 a"
|
a ) 273054 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273258
|
a
|
multiply(divide(5354, 51), const_100)
|
divide(n0,n1)|multiply(#0,const_100)|
|
general
|
5354 x 51 = ?
|
we write n ! = n * ( n - 1 ) ( n - 2 ! ) therefore n ! / ( n - 2 ) ! = n ( n - 1 ) * ( n - 2 ) ! / ( n - 2 ) ! = n ( n - 1 ) . - - > n ( n - 1 ) = 342 - - > n ^ 2 - n - 342 = 0 - - > n ^ 2 - 19 n + 18 n - 342 = 0 - - > n ( n - 19 ) + 18 ( n - 19 ) = 0 - - > ( n - 19 ) ( n + 18 ) = 0 therefore n - 19 = 0 ; n + 18 = 0 ; ( i . e ) n = 19 ; n = - 18 we want positive integer . so , n = 19 . answer : c
|
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21
|
c
|
sqrt(add(342, divide(const_1, const_4)))
|
divide(const_1,const_4)|add(n1,#0)|sqrt(#1)
|
general
|
if n is a positive integer such that n ! / ( n - 2 ) ! = 342 , find n .
|
let amount of 10 $ be x . then amount of 5 $ be 2 x . now 5 * 2 x + 10 * x = 100 . thus x = 5 . answer : e
|
a ) 2 , b ) 6 , c ) 7 , d ) 8 , e ) 5
|
e
|
divide(divide(100, 10), const_2)
|
divide(n0,n1)|divide(#0,const_2)
|
general
|
a person has 100 $ in 10 $ and 5 $ bill . if the 5 $ bill quantity is twice that of 10 $ bill . what is quantity of 10 $ .
|
"index for females = ( 20 - 5 ) / 20 = 3 / 4 = 0.75 index for males = ( 20 - 15 / 20 = 1 / 4 = 0.25 index for females exceeds males by 0.75 - 0.25 = 0.5 answer : c"
|
a ) 0.05 , b ) 0.0625 , c ) 0.5 , d ) 0.25 , e ) 0.6
|
c
|
subtract(divide(subtract(20, 5), 20), divide(5, 20))
|
divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)|
|
general
|
for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 5 of whom are females , by how much does the index for the females exceed the index for the males in the group ?
|
"the radius of tank a is 12 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 144 / ( 4 * pi ^ 2 ) = 360 / ( pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 12 * pi * 100 / ( 4 * pi ^ 2 ) = 300 / ( pi ) tank a / tank b = 360 / 300 = 12 / 10 = 120 % the answer is e ."
|
a ) 80 % , b ) 90 % , c ) 100 % , d ) 110 % , e ) 120 %
|
e
|
multiply(multiply(power(divide(12, 10), const_2), divide(10, 12)), const_100)
|
divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)|
|
physics
|
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 12 meters , and the interior of tank b has a height of 12 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ?
|
sandra score can be like 1,4 , 9,16 , 25,36 eric score less then 1 - - > 0 eric score less then 4 = ( 1,1 ) , ( 1,2 ) ( 2,1 ) - - > 3 eric score less then 9 are ( 1,1 ) ( 1,2 ) ( 1,3 ) ( 1,4 ) ( 1,5 ) ( 1,6 ) ( 2,1 ) ( 2,2 ) ( 2,3 ) ( 2,4 ) ( 2,5 ) ( 2,6 ) ( 3,1 ) ( 3,2 ) ( 3,3 ) ( 3,4 ) ( 3,5 ) ( 4,1 ) ( 4,2 ) ( 4,3 ) ( 4,4 ) ( 5,1 ) ( 5,2 ) ( 5,3 ) ( 6,1 ) ( 6,2 ) - - > 26 eric score will always be less then 16 - - - > 36 eric score will always be less then 25 - - - > 36 eric score will always be less then 36 - - - > 36 total favorable outcomes = 3 + 26 + 36 + 36 + 36 = 137 total possible outcomes = 216 ( 36 * 6 ) probability = 137 / 216 answer : a
|
['a ) 137 / 216', 'b ) 137 / 218', 'c ) 137 / 217', 'd ) 136 / 216', 'e ) 138 / 216']
|
a
|
divide(add(add(add(add(const_3, subtract(power(multiply(2, const_3), const_2), const_10)), power(multiply(2, const_3), const_2)), power(multiply(2, const_3), const_2)), power(multiply(2, const_3), const_2)), multiply(power(multiply(2, const_3), const_2), multiply(2, const_3)))
|
multiply(n0,const_3)|power(#0,const_2)|multiply(#0,#1)|subtract(#1,const_10)|add(#3,const_3)|add(#4,#1)|add(#5,#1)|add(#6,#1)|divide(#7,#2)
|
geometry
|
eric throws 2 dice , and his score is the sum of the values shown . sandra throws one dice and her score is the square of the value shown . what is the probabilty that sandras score will be strictly higher than erics score ? ?
|
worker b will get 9 / 14 = 64.29 % the answer is c .
|
a ) 62.27 % , b ) 63.28 % , c ) 64.29 % , d ) 65.31 % , e ) 66.32 %
|
c
|
divide(910, add(5, 9))
|
add(n1,n2)|divide(n0,#0)
|
other
|
if $ 910 are divided between worker a and worker b in the ratio 5 : 9 , what is the share that worker b will get ?
|
"per second = > 8 * 5280 ft / 60 * 60 = 11.73 ft 5 seconds = > 11.73 * 5 = 58.65 ft answer : e"
|
a ) 60 ft , b ) 52 ft , c ) 53 ft , d ) 55 ft , e ) 58.65 ft
|
e
|
multiply(5, divide(multiply(8, 5280), const_3600))
|
multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)|
|
physics
|
someone on a skateboard is traveling 8 miles per hour . how many feet does she travel in 5 seconds ? ( 1 mile = 5280 feet )
|
"30 * t = 1 km = > t = 1 / 30 km / h v * ( t + 30 / 3600 ) = 1 v ( 1 / 30 + 30 / 3600 ) = 1 v ( 150 / 3600 ) = 1 v = 24 km / h the answer is e ."
|
a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24
|
e
|
divide(1, divide(add(multiply(const_3600, divide(1, 30)), 30), const_3600))
|
divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)|
|
physics
|
a car traveling at a certain constant speed takes 30 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 30 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ?
|
say a 100 families existed in 1998 then the number of families owning a computer in 1998 - 30 number of families owning computer in 2002 = 30 * 120 / 100 = 36 number of families in 2002 = 108 the percentage = 36 / 108 * 100 = 33.33 % . option : d
|
a ) 50.12 % , b ) 52.66 % , c ) 56.33 % , d ) 33.33 % , e ) 74.12 %
|
d
|
multiply(const_100, divide(divide(multiply(add(20, const_100), 30), const_100), add(const_100, 8)))
|
add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)
|
general
|
of the families in city x in 1998 , 30 percent owned a personal computer . the number of families in city x owning a computer in 2002 was 20 percent greater than it was in 1998 , and the total number of families in city x was 8 percent greater in 2002 than it was in 1998 . what percent of the families in city x owned a personal computer in 2002 ?
|
26 steps 30 seconds and for 34 steps only 18 seconds left to reach botto . means he covered 8 steps ( i . e . 34 - 26 ) in 12 ( i . e 30 - 18 ) seconds the spped of the boy is 8 steps in 12 seconds after further simplyfy . . 2 steps in 3 seconds after 34 steps only 18 seconds , means 12 more steps are left total steps are 34 + 12 = 46 answer : e
|
a ) 43 , b ) 44 , c ) 45 , d ) 40 , e ) 46
|
e
|
add(add(multiply(const_3, const_10), 4), multiply(divide(subtract(add(multiply(const_3, const_10), 4), add(20, 6)), subtract(multiply(const_3, const_10), multiply(6, const_3))), multiply(6, const_3)))
|
add(n0,n1)|multiply(const_10,const_3)|multiply(n1,const_3)|add(n2,#1)|subtract(#1,#2)|subtract(#3,#0)|divide(#5,#4)|multiply(#6,#2)|add(#3,#7)
|
physics
|
recently , i decided to walk down an escalator of a tube station . i did some quick calculation in my mind . i found that if i walk down 20 ` ` 6 steps , i require thirty seconds to reach the bottom . however , if i am able to step down thirty ` ` 4 stairs , i would only require eighteen seconds to get to the bottom . if the time is measured from the moment the top step begins to descend to the time i step off the last step at the bottom ?
|
"soln : - can be reduced to y / 10 + 3 y / 10 = 2 y / 5 = 40 % answer : a"
|
a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 %
|
a
|
multiply(const_100, add(divide(2, 20), divide(3, 10)))
|
divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)|
|
general
|
if y > 0 , ( 2 y ) / 20 + ( 3 y ) / 10 is what percent of y ?
|
let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 5 x / 6 ) ( 2 y / 3 ) = 5 xy / 9 the total number of boxes is xy + 5 xy / 9 = 14 xy / 9 the fraction loaded by the day crew is xy / ( 14 xy / 9 ) = 9 / 14 the answer is d .
|
a ) 6 / 11 , b ) 7 / 12 , c ) 8 / 13 , d ) 9 / 14 , e ) 11 / 15
|
d
|
divide(multiply(6, 3), add(multiply(6, 3), multiply(2, 5)))
|
multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)
|
physics
|
at a loading dock , each worker on the night crew loaded 2 / 3 as many boxes as each worker on the day crew . if the night crew has 5 / 6 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ?
|
"e e = 90 * 5 / 18 * 6 = 150 m"
|
a ) 281 m , b ) 112 m , c ) 117 m , d ) 125 m , e ) 150 m
|
e
|
multiply(multiply(90, const_0_2778), 6)
|
multiply(n0,const_0_2778)|multiply(n1,#0)|
|
physics
|
if a train , travelling at a speed of 90 kmph , crosses a pole in 6 sec , then the length of train is ?
|
explanation : l . c . m of 25650 = 2 x 3 x 3 x 3 x 5 x 5 x 19 3 , 2 , 5,19 number of different prime factors is 4 . answer : option a
|
a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6
|
a
|
add(const_2, const_2)
|
add(const_2,const_2)
|
other
|
find the number of different prime factors of 25650
|
"correct sum = ( 36 * 50 + 58 - 43 ) = 1815 . correct mean = 1815 / 50 = 36.3 answer : e"
|
a ) 36.7 , b ) 36.1 , c ) 36.5 , d ) 36.9 , e ) 36.3
|
e
|
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 43)), 50)
|
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
|
general
|
the mean of 50 observations was 36 . it was found later that an observation 58 was wrongly taken as 43 . the corrected new mean is ?
|
"44 + 28 + 10 = 82 % 100 – 82 = 18 % 450 * 18 / 100 = 81 answer : d"
|
a ) 72 , b ) 75 , c ) 80 , d ) 81 , e ) 90
|
d
|
divide(multiply(450, subtract(const_100, add(add(44, 28), 10))), const_100)
|
add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)|
|
gain
|
in a school of 450 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?
|
"oa is ' c ' . oe : take the remainder from each of 1250 / 32 , 1040 / 32 and so on . . 1250 / 32 gives remainder = 2 1040 / 32 gives remainder = 16 1057 / 32 gives remainder = 1 1145 / 32 gives remainder = 25 the net remainder is the product of above individual remainders . i . e = 2 * 16 * 1 * 25 break them into pairs 2 * 16 / 32 gives remainder 0 and 1 * 25 / 32 gives remainder 25 so 0 * 25 / 32 gives remainder 0 . answer : c"
|
a ) 3 , b ) 5 , c ) 0 , d ) 2 , e ) 9
|
c
|
reminder(multiply(1040, 1250), 1057)
|
multiply(n0,n1)|reminder(#0,n2)|
|
general
|
what is the remainder when 1250 * 1040 * 1057 * 1145 is divided by 32 ?
|
"let c . p . of each article be re . 1 . then , c . p . of 80 articles = rs . 80 ; s . p . of 80 articles = rs . 40 . loss % = 40 / 80 * 100 = 50 % answer : c"
|
a ) 30 % , b ) 35 % , c ) 50 % , d ) 55 % , e ) 40 %
|
c
|
subtract(80, 40)
|
subtract(n0,n1)|
|
gain
|
if the selling price of 80 articles is equal to the cost price of 40 articles , then the loss or gain percent is :
|
volume = ( 16 x 14 x 7 ) m ^ 3 = 1568 m ^ 3 . surface area = [ 2 ( 16 x 14 + 14 x 7 + 16 x 7 ) ] cm ^ 2 = ( 2 x 434 ) cm ^ 2 = 868 cm ^ 2 . answer is d
|
['a ) 878 cm ^ 2', 'b ) 858 cm ^ 2', 'c ) 838 cm ^ 2', 'd ) 868 cm ^ 2', 'e ) none of them']
|
d
|
multiply(add(multiply(16, 7), add(multiply(16, 14), multiply(14, 7))), const_2)
|
multiply(n0,n1)|multiply(n1,n2)|multiply(n0,n2)|add(#0,#1)|add(#3,#2)|multiply(#4,const_2)
|
geometry
|
find the volume and surface area of a cuboid 16 m long , 14 m broad and 7 m high .
|
"sol . income in first year = * x 42000 = rs . 24000 expenses in second year = \ x 21000 = rs . 35000 total savings = total income - total expenses = ( 42000 + 24000 ) - ( 24000 + 35000 ) = 66000 - 59000 = rs . 7000 b"
|
a ) s . 8000 , b ) s . 7000 , c ) s . 9900 , d ) s . 9990 , e ) s . 10000
|
b
|
add(subtract(42000, divide(multiply(24000, 5), 3)), subtract(divide(multiply(42000, 4), 7), 24000))
|
multiply(n4,n7)|multiply(n1,n6)|divide(#0,n3)|divide(#1,n2)|subtract(n6,#2)|subtract(#3,n7)|add(#4,#5)|
|
general
|
for 2 consecutive yrs , my incomes are in the ratio of 4 : 7 and expenses in the ratio of 3 : 5 . if my income in the 2 nd yr is rs . 42000 & my expenses in the first yr in rs . 24000 , my total savings for the two - year is
|
{ total } = { retirees } + { bridge players } - { both } + { neither } x = 5 / 20 * x + x / 2 - 3 / 20 * x + 120 20 x = 5 x + 10 x - 3 x + 120 * 20 ( multiply by 20 ) 12 x = 120 * 20 x = 200 . answer : c
|
a ) 240 , b ) 300 , c ) 200 , d ) 400 , e ) 480
|
c
|
multiply(subtract(120, multiply(5, const_4)), const_2)
|
multiply(n1,const_4)|subtract(n2,#0)|multiply(#1,const_2)
|
general
|
3 - twentieths of the members of a social club are retirees who are also bridge players , 5 - twentieths of the members are retirees , and one - half of the members are bridge players . if 120 of the members are neither retirees nor bridge players , what is the total number of members in the social club ?
|
"description : = > x + y = 8 , x - y = 6 adding these 2 x = 14 = > x = 7 , y = 1 . thus the number is 71 answer d"
|
a ) 85 , b ) 94 , c ) 83 , d ) 71 , e ) none
|
d
|
add(multiply(divide(add(8, 6), const_2), 8), subtract(8, divide(add(8, 6), const_2)))
|
add(n0,n1)|divide(#0,const_2)|multiply(n0,#1)|subtract(n0,#1)|add(#2,#3)|
|
general
|
the sum of digits of a two digit number is 8 , the difference between the digits is 6 . find the number
|
let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 80 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 80 + 20 ) / 2 = 50 . answer : d
|
a ) 40 , b ) 30 , c ) 25 , d ) 50 , e ) none of these .
|
d
|
divide(add(80, 20), const_2)
|
add(n0,n1)|divide(#0,const_2)
|
general
|
the total marks obtained by a student in mathematics and physics is 80 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .
|
"area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 44 / 2 = 55 cm 2 answer : b"
|
a ) 38 cm 2 , b ) 55 cm 2 , c ) 65 cm 2 , d ) 45 cm 2 , e ) 35 cm 2
|
b
|
triangle_area(2.5, 44)
|
triangle_area(n0,n1)|
|
geometry
|
the perimeter of a triangle is 44 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ?
|
"principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 4903 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 4076 . answer : b"
|
a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096
|
b
|
divide(4903, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3))
|
divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)|
|
gain
|
the principal that amounts to rs . 4903 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ?
|
"number of students taking history = h = 36 number of students taking statistics = s = 32 total number of students = t = 90 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 95 - 59 = 36 letxbe the number of students taking both history and statistics . then t = h + s + n - x or 95 = 36 + 32 + 36 - x or x = 9 now , number of students taking only history will be h - x or 36 - 9 = 27 answer : - e"
|
a ) 9 , b ) 19 , c ) 23 , d ) 45 , e ) 27
|
e
|
subtract(36, subtract(add(36, 32), 59))
|
add(n1,n2)|subtract(#0,n3)|subtract(n1,#1)|
|
other
|
in a group of 95 students , 36 are taking history , and 32 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ?
|
"the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 80 = $ 1.00 cents on commodity y , which will take 10 years . the answer is b ."
|
a ) 2010 . , b ) 2011 . , c ) 2012 . , d ) 2013 . , e ) 2014 .
|
b
|
add(2001, divide(add(divide(80, const_100), subtract(4.40, 4.20)), subtract(divide(30, const_100), subtract(4.40, 4.20))))
|
divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)|
|
general
|
the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 80 cents more than the commodity y ?
|
favorable case = the club is picked in the third draw or later unfavorable cases = the club is picked in either first draw , second draw or third draws probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case 1 : probability of club picked in first draw = 13 / 52 = 1 / 4 unfavorable case 2 : probability of club picked in second draw ( i . e . first draw is not club ) = ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 1 / 4 ) = 3 / 16 unfavorable case 3 : probability of club picked in third draw ( i . e . first and 2 nd draws are not clubs ) = ( 39 / 52 ) * ( 39 / 52 ) * ( 13 / 52 ) = ( 3 / 4 ) * ( 3 / 4 ) * ( 1 / 4 ) = 9 / 64 total unfavorable probability = ( 1 / 4 ) + ( 3 / 16 ) + ( 9 / 64 ) = ( 16 / 64 ) + ( 12 / 64 ) + ( 9 / 64 ) = 37 / 64 i . e . , favorable probability = 1 - ( 37 / 64 ) = 27 / 64 answer : option e
|
a ) 1 / 2 , b ) 3 / 4 , c ) 7 / 8 , d ) 27 / 32 , e ) 27 / 64
|
e
|
multiply(multiply(divide(subtract(52, 13), 52), divide(subtract(52, 13), 52)), divide(subtract(52, 13), 52))
|
subtract(n1,n0)|divide(#0,n1)|multiply(#1,#1)|multiply(#1,#2)
|
probability
|
there are 13 clubs in a full deck of 52 cards . in a certain game , you pick a card from a standard deck of 52 cards . if the card is a club , you win . if the card is not a club , the person replaces the card to the deck , reshuffles , and draws again . the person keeps repeating that process until he picks a club , and the point is to measure how many draws it took before the person picked a club and , thereby , won . what is the probability that one will pick the first club on the forth draw or later ?
|
"a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 80 = 25 answer : b"
|
a ) 15 , b ) 25 , c ) 35 , d ) 45 , e ) 55
|
b
|
subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 70))
|
add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)|
|
general
|
the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 70 , and what number ?
|
"assume the cost of the least sized ( 26 inch ) tv = x cost of 28 inches tv = x + 200 cost of 30 inches tv = x + 400 cost of 32 inches tv = x + 600 total cost = 4 x + 1200 = 3200 therefore x = 2000 / 4 = 500 price of 28 inch + 30 inch = 700 + 900 = 1600 option b"
|
a ) 850 , b ) 1,600 , c ) 1,700 , d ) 2,550 , e ) 3,400
|
b
|
divide(add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), 200)), subtract(multiply(const_2, const_10), const_1))
|
add(n5,const_3600)|add(const_1,const_2)|multiply(const_10,const_2)|add(#1,const_3)|subtract(#2,const_1)|multiply(n5,#3)|subtract(#0,#5)|divide(#6,n0)|add(n5,#7)|add(n5,#8)|add(#8,#9)|divide(#10,#4)|
|
general
|
alan buys 4 tvs , a 26 inch , a 28 inch , a 30 inch , and a 32 inch , for his new house . each tv costs $ 200 more than the size below it . alan spends a total of $ 3,200 . how much would he have spent if he had bought only the 28 inch and 30 inch tvs ?
|
use the elimination method to find the correct option . of all the options only 90 fits 28 % of 90 = 25.2 18 % of 90 = 16.2 25.2 - 16.2 = 7.2 required number is 90 . answer : e
|
a ) 50 , b ) 34 , c ) 55 , d ) 60 , e ) 90
|
e
|
add(divide(7.2, divide(subtract(28, 18), const_100)), 18)
|
subtract(n0,n1)|divide(#0,const_100)|divide(n2,#1)|add(n1,#2)
|
gain
|
if 28 % of a number exceeds 18 % of it by 7.2 , then find the number ?
|
prime factors of 60 are 2 ^ 2,3 ^ 1,5 ^ 1 total divisors = ( power if a prime factor + 1 ) total no . of odd factors ( 3,5 , ) = ( 1 + 1 ) ( 1 + 1 ) = 4 since we need odd divisors other than 1 = > 4 - 1 = 3 odd divisors d is the answer
|
a ) 3 , b ) 4 , c ) 5 , d ) 3 , e ) 7
|
d
|
divide(60, multiply(const_10, const_2))
|
multiply(const_10,const_2)|divide(n0,#0)
|
other
|
how many factors of 60 are odd numbers greater than 1 ?
|
d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , e = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d . d
|
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
|
d
|
subtract(const_100, divide(subtract(const_100, 68), const_2))
|
subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)
|
general
|
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent e of the distribution is less than m + d ?
|
p = q + 1150 . p = 4 ( q - 200 ) . 4 ( q - 200 ) = q + 1150 . 3 q = 1950 . q = 650 . the answer is c .
|
a ) 600 , b ) 625 , c ) 650 , d ) 675 , e ) 700
|
c
|
divide(add(1150, multiply(200, 4)), const_3)
|
multiply(n1,n2)|add(n0,#0)|divide(#1,const_3)
|
general
|
village p ’ s population is 1150 greater than village q ' s population . if village q ’ s population were reduced by 200 people , then village p ’ s population would be 4 times as large as village q ' s population . what is village q ' s current population ?
|
"explanation : we can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak , as work done for 1 hour without leak = 1 / 3 work done with leak = 3 1 / 2 = 7 / 2 work done with leak in 1 hr = 2 / 7 work done by leak in 1 hr = 1 / 3 = 2 / 7 = 1 / 21 so tank will be empty by the leak in 21 hours . answer is d"
|
a ) 10 hours , b ) 13 hours , c ) 17 hours , d ) 21 hours , e ) 25 hours
|
d
|
divide(3, const_1)
|
divide(n1,const_1)|
|
physics
|
an electric pump can fill a tank in 3 hours . because of a leak in the tank , it took 3 hours 30 min to fill the tank . in what time the leak can drain out all the water of the tank and will make tank empty ?
|
"now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 35 % profit . this means the clocks ' selling price was c ( 1.35 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 15 % the price at which the collector bought it . thus , you get = 1.35 * 0.15 * c = 0.2025 c furthermore , the store sells the clock for the second time for 65 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.65 = 1.65 * 0.2025 c finally given that c - 0.2025 c = 105 - - - - > c = 131.66 $ thus , the cost of the clock the second time around = 1.65 * 0.2025 c = 1.65 * 0.2025 * 131.66 = 43.99 $ . hence e is the correct answer ."
|
a ) $ 75.44 , b ) $ 88.23 , c ) $ 64.77 , d ) $ 54.32 , e ) $ 43.99
|
e
|
divide(multiply(105, divide(multiply(add(105, 65), divide(add(105, 35), const_2)), 105)), subtract(divide(add(105, 35), const_2), 35))
|
add(n2,n3)|add(n0,n3)|divide(#1,const_2)|multiply(#0,#2)|subtract(#2,n0)|divide(#3,n3)|multiply(n3,#5)|divide(#6,#4)|
|
general
|
a clock store sold a certain clock to a collector for 35 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 15 percent of what the collector had paid . the shop then sold the clock again at a profit of 65 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 105 , for how much did the shop sell the clock the second time ?
|
"present age of total members = 6 x 26 = 156 10 yrs back their ages were = 6 x 10 = 60 ages at the birth of youngest member = 156 - 60 = 96 therefore , avg age at the birth of youngest member = 96 / 6 = 16 . answer : c"
|
a ) 15 , b ) 18 , c ) 16 , d ) 12 , e ) 19
|
c
|
subtract(divide(subtract(multiply(6, 26), 10), const_4), 10)
|
multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_4)|subtract(#2,n2)|
|
general
|
the average age of a family of 6 members is 26 years . if the age of the youngest member is 10 years , what was the average age of the family at the birth of the youngest member ?
|
say the rates of machines a , b and c are a , b , and c , respectively . together 15 type a machines and 7 type b machines can complete a certain job in 4 hours - - > 15 a + 7 b = 1 / 4 ; together 8 type b machines and 15 type c machines can complete the same job in 11 hours - - > 8 b + 15 c = 1 / 11 . sum the above : 15 a + 15 b + 15 c = 1 / 4 + 1 / 11 = 15 / 44 - - > reduce by 15 : a + b + c = 1 / 44 - - > so , the combined rate of the three machines is 1 / 44 job / hour - - > time is reciprocal of the rate , thus machines a , b and c can do the job e in 44 hours . answer : c .
|
a ) 22 hours , b ) 30 hours , c ) 44 hours , d ) 60 hours , e ) it can not be determined from the information above .
|
c
|
divide(const_1, divide(add(divide(const_1, 4), divide(const_1, 11)), 15))
|
divide(const_1,n2)|divide(const_1,n5)|add(#0,#1)|divide(#2,n0)|divide(const_1,#3)
|
physics
|
together , 15 type a machines and 7 type b machines can complete a certain job in 4 hours . together 8 type b machines and 15 type c machines can complete the same job in 11 hours . how many hours e would it take one type a machine , one type b machine , and one type c machine working together to complete the job ( assuming constant rates for each machine ) ?
|
"1 = 5,2 = 4,3 = 253,4 = 2,6 = 15 , then 15 = ? 15 = 6 check the fifth eqn . answer : e"
|
a ) 1 , b ) 255 , c ) 345 , d ) 445 , e ) 6
|
e
|
divide(subtract(subtract(15, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2)
|
add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)|
|
general
|
1 = 5,2 = 4,3 = 253,4 = 2,6 = 15 , then 15 = ?
|
"the pattern is 1 × 2 , 2 × 3 , 3 × 4 , 4 × 5 , 5 × 6 , 6 × 7 , 7 × 8 . hence , next number is 8 × 9 = 72 answer is c"
|
a ) 62 , b ) 74 , c ) 72 , d ) 81 , e ) 58
|
c
|
subtract(negate(20), multiply(subtract(6, 12), divide(subtract(6, 12), subtract(2, 6))))
|
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
|
general
|
2 , 6 , 12 , 20 , 30 , 42 , 56 , ( . . . )
|
"16 * 16 = 256 sq m the answer is d ."
|
a ) 225 sq m , b ) 286 sq m , c ) 298 sq m , d ) 256 sq m , e ) 231 sq m
|
d
|
divide(square_area(16), const_2)
|
square_area(n0)|divide(#0,const_2)|
|
geometry
|
what is the area of a square field whose sides have a length of 16 meters ?
|
"30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 50,000 i . e . rs . 25,000 answer : c"
|
a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 25,000 , d ) rs . 36,000 , e ) none of these
|
c
|
multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3)))
|
add(n1,const_3)|multiply(n1,const_4)|multiply(const_3,const_4)|multiply(#0,n1)|multiply(#0,const_100)|divide(#3,n1)|multiply(#1,#3)|multiply(#4,const_100)|divide(#5,#2)|multiply(#6,const_100)|add(#9,#7)|multiply(#10,#8)|
|
gain
|
jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 50,000 after completion of one year . what will be madhu ' s share of profit ?
|
"solution ( place value of 7 ) - ( place value of 3 ) = ( 7000 - 30 ) = 6970 . answer d"
|
a ) 4 , b ) 5 , c ) 45 , d ) 6970 , e ) none
|
d
|
subtract(multiply(const_10, 7), 7)
|
multiply(n0,const_10)|subtract(#0,n0)|
|
general
|
the difference between the place values of 7 and 3 in the prime number 527435 is
|
"the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 160 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 160 . so 1 part corresponds to 160 / 8 = 20 and so 5 parts correspond to 5 * 10 = 100 . a"
|
a ) 100 , b ) 36 , c ) 45 , d ) 72 , e ) 117
|
a
|
subtract(divide(160, subtract(const_1, divide(5, 13))), 160)
|
divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|subtract(#2,n2)|
|
other
|
in a certain school , the ratio of boys to girls is 5 to 13 . if there are 160 more girls than boys , how many boys are there ?
|
"1 hour he walk 4 km he walk 40 km in = 40 / 4 * 1 = 10 hours answer is b"
|
a ) 5 hrs , b ) 10 hrs , c ) 15 hrs , d ) 20 hrs , e ) 30 hrs
|
b
|
divide(40, 4)
|
divide(n2,n0)|
|
physics
|
ajay can walk 4 km in 1 hour . in how many hours he can walk 40 km ?
|
if ramesh attempts ' x ' questions correct and ' y ' questions wrong , then x + y = 108 - - - ( i ) & x - ( 1 / 3 ) y = 0 - - - ( ii ) on solving x = 27 , y = 81 answer : d
|
a ) 78 , b ) 79 , c ) 80 , d ) 81 , e ) 82
|
d
|
subtract(108, divide(multiply(divide(1, 3), 108), add(const_1, divide(1, 3))))
|
divide(n2,n3)|add(#0,const_1)|multiply(n0,#0)|divide(#2,#1)|subtract(n0,#3)
|
general
|
ramesh has solved 108 questions in an examination . if he got only ‘ 0 ’ marks , then how many questions were wrong when one mark is given for each one correct answer and 1 / 3 mark is subtracted on each wrong answer .
|
"c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be x = 60 . c"
|
a ) 30 , b ) 45 , c ) 60 , d ) 75 , e ) 90
|
c
|
divide(const_180, const_3)
|
divide(const_180,const_3)|
|
geometry
|
what is the measure of the angle x made by the diagonals of the any adjacent sides of a cube .
|
"cp * ( 85 / 100 ) = 915 cp = 10.76 * 100 = > cp = 1076 answer : c"
|
a ) 915 , b ) 1000 , c ) 1076 , d ) 1067 , e ) 1760
|
c
|
divide(915, subtract(const_1, divide(15, const_100)))
|
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
|
gain
|
after decreasing 15 % in the price of an article costs rs . 915 . find the actual cost of an article ?
|
"dan can complete 1 / 15 of the job per hour . in 12 hours , dan completes 12 ( 1 / 15 ) = 4 / 5 of the job . annie can complete 1 / 10 of the job per hour . to complete the job , annie will take 1 / 5 / 1 / 10 = 2 hours . the answer is a ."
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
|
a
|
multiply(subtract(const_1, divide(12, 15)), 10)
|
divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)|
|
physics
|
dan can do a job alone in 15 hours . annie , working alone , can do the same job in just 10 hours . if dan works alone for 12 hours and then stops , how many hours will it take annie , working alone , to complete the job ?
|
here all percentage when summed we need to get 100 % . as per data 17 + 24 + 15 + 29 + 8 = 93 % . so remaining 7 % is the balance for the astrophysics . since this is a circle all percentage must be equal to 360 degrees . 100 % - - - - 360 degrees then 7 % will be 26 degrees . . imo option c .
|
a ) 8 ° , b ) 10 ° , c ) 26 ° , d ) 36 ° , e ) 52 °
|
c
|
divide(multiply(subtract(const_100, add(add(add(add(17, 24), 15), 29), 8)), divide(const_3600, const_10)), const_100)
|
add(n0,n1)|divide(const_3600,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|subtract(const_100,#4)|multiply(#1,#5)|divide(#6,const_100)
|
gain
|
a circle graph shows how the megatech corporation allocates its research and development budget : 17 % microphotonics ; 24 % home electronics ; 15 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ?
|
"a = 1600 ( 21 / 20 ) 3 = 2522 answer : c"
|
a ) 3388 , b ) 2778 , c ) 2522 , d ) 2988 , e ) 2771
|
c
|
subtract(multiply(1600, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 1600)
|
divide(const_0_25,const_4)|add(#0,const_1)|multiply(#1,#1)|multiply(n0,#2)|subtract(#3,n0)|
|
gain
|
find the c . i . on a sum of rs . 1600 for 9 months at 20 % per annum , interest being compounded quarterly ?
|
"explanation : 20 cp = 25 sp cost price cp = 25 selling price sp = 20 formula = ( sp - cp ) / cp * 100 = ( 20 - 25 ) / 25 * 100 = 20 % loss answer : option a"
|
a ) 20 % loss , b ) 30 % loss , c ) 40 % gain , d ) 40 % loss , e ) 50 % loss
|
a
|
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 20)), divide(multiply(const_100, 25), 20)))
|
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
|
gain
|
if the cost price of 20 articles is same as the selling price of 25 articles . find the gain or loss percentage ?
|
"since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 830 = 24 interest earned for 3 years = 24 * 3 = 72 amount invested = 830 - 72 = 758 answer : d"
|
a ) 670 , b ) 664 , c ) 698 , d ) 758 , e ) 700
|
d
|
subtract(830, multiply(divide(subtract(854, 830), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100)))
|
divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5)|
|
gain
|
peter invests a sum of money and gets back an amount of $ 830 in 3 years . david invests an equal amount of money and gets an amount of $ 854 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ?
|
let the amount be x . from the question , x × 14 × 6 / 1200 − x × 8 × 8 / 1200 = 250 ∴ x = 15000 answer a
|
a ) 15000 , b ) 25000 , c ) 7500 , d ) 14500 , e ) none of these
|
a
|
divide(250, subtract(multiply(divide(14, const_12), divide(6, const_100)), multiply(divide(8, const_12), divide(8, const_100))))
|
divide(n0,const_12)|divide(n1,const_100)|divide(n3,const_12)|divide(n3,const_100)|multiply(#0,#1)|multiply(#2,#3)|subtract(#4,#5)|divide(n2,#6)
|
general
|
the simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the rate of 8 per cent in 8 months . how much amount was borrowed ?
|
"there is no certain way to solve 2 unknown with 1 equation . the best way is to look at the question and retrospect the most efficient way . in this question , a and b are only positive integers . so that is a big relief . now , we can start with putting a = 1,2 , . . and so on till the time we are confident about one of the options . so , we start with a = 1 , we get b as - ve . out a = 2 , we get b as 6 . yes ( now ( a , b ) = ( 2,6 ) . we can directly see that ( a , b ) = ( 6,2 ) will also satisfy . so we have 2 possible solutions ) a = 3 , we get b as 3 . yes ( now we have 3 possible solutions ) a = 4 , we get b as fraction . out a = 5 , we get b again as some fraction . out a = 6 already taken . we have a , b options left . c , d , e are out . a is 6 . to have 6 as the answer , we will need one more pair like 2,6 and one more solution where a = b . when a = b , we have only 1 solution = 5 . so , one more solution , where a = b is not possible . so , answer will be e ."
|
a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 5
|
e
|
divide(34, 57)
|
divide(n2,n3)|
|
general
|
how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 34 / 57 ?
|
"120 amounts to 240 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 240 120 + 120 * ( r / 100 ) * ( 3 ) = 240 = > r = 100 / 3 150 in 6 years = principal + interest = 150 + 150 * ( r / 100 ) * ( 6 ) = 450 answer is e ."
|
a ) $ 190 , b ) $ 180 , c ) $ 200 , d ) $ 340 , e ) $ 450
|
e
|
add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(240, 120), 120), 120), 3)), 120))
|
multiply(n3,n4)|subtract(n1,n0)|multiply(#1,n0)|divide(#2,n0)|divide(#3,n2)|multiply(#4,#0)|divide(#5,n0)|add(n3,#6)|
|
gain
|
if $ 120 invested at a certain rate of simple interest amounts to $ 240 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ?
|
"total manufacturing cost = 10500 + 1000 * 30 = 40500 total selling cost = 1000 * 60 = 60000 profit = 60000 - 40500 = 19500 answer : a"
|
a ) $ 19500 , b ) $ 15500 , c ) $ 29500 , d ) $ 39500 , e ) $ 1500
|
a
|
multiply(subtract(60, multiply(30, const_2)), 1000)
|
multiply(n1,const_2)|subtract(n2,#0)|multiply(n3,#1)|
|
gain
|
redo ’ s manufacturing costs for sets of horseshoes include a $ 10,500 initial outlay , and $ 30 per set . they can sell the sets $ 60 . if profit is revenue from sales minus manufacturing costs , and the company producessells 1000 sets of horseshoes , what was their profit ?
|
we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 6 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : 9 x + 6 x + 5 x = 40 20 x = 40 x = 2 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( 5 ) ( 2 ) = 10 pounds of apples = ( 9 ) ( 2 ) = 18 thus we know that there are 18 – 10 = 8 more pounds of apples than grapes . answer is c .
|
a ) 15 , b ) 12 , c ) 8 , d ) 6 , e ) 4
|
c
|
subtract(multiply(9, const_2), multiply(5, const_2))
|
multiply(n0,const_2)|multiply(n2,const_2)|subtract(#0,#1)
|
general
|
a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 9 : 6 : 5 , respectively , by weight . if 40 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ?
|
"explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1500 * 100 / 120 * 100 / 125 = 1000 answer : option a"
|
a ) a ) 1000 , b ) b ) 1100 , c ) c ) 1200 , d ) d ) 1300 , e ) e ) 1400
|
a
|
divide(1500, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))
|
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|
|
gain
|
the population of a town increases 20 % and 25 % respectively in two consecutive years . after the growth the present population of the town is 1500 . then what is the population of the town 2 years ago ?
|
"one case is : 1 / 4 * 1 / 4 * 3 / 4 * 3 / 4 = 9 / 256 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 9 / 256 ) = 27 / 128 the answer is b ."
|
a ) 14 / 65 , b ) 27 / 128 , c ) 33 / 140 , d ) 37 / 165 , e ) 41 / 187
|
b
|
subtract(1, divide(const_2, 4))
|
divide(const_2,n1)|subtract(n0,#0)|
|
general
|
when a random experiment is conducted , the probability that event a occurs is 1 / 4 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ?
|
"64 / 5000 = 64 / ( 5 * 10 ^ 3 ) = ( 64 / 5 ) * 10 ^ - 3 = 12.8 * 10 ^ - 3 = . 0128 thousandths digit = 2 answer a"
|
a ) 2 , b ) 1 , c ) 3 , d ) 5 , e ) 6
|
a
|
floor(multiply(const_100, divide(64, 5000)))
|
divide(n0,n1)|multiply(#0,const_100)|floor(#1)|
|
general
|
what is the thousandths digit in the decimal equivalent of 64 / 5000 ?
|
"solution average = ( 10 + 15 + 20 + 25 + 30 / 5 ) = 100 / 5 = 20 . answer b"
|
a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 32
|
b
|
divide(add(add(6, const_4), subtract(34, const_4)), const_2)
|
add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|
|
general
|
find the average of all the numbers between 6 and 34 which are divisible by 5 .
|
"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 24 cents ; a half - liter of kerosene costs as much as 8 eggs - - > 8 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 16 eggs = 16 / 12 * 24 = 32 cents . answer : c ."
|
a ) 0.32 , b ) 0.44 , c ) 32 , d ) 44 , e ) 55
|
c
|
multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.24, 100))
|
divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|
|
general
|
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