Problem
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Rationale
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300
correct
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annotated_formula
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in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation ....
"standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 1 . so , x = 65 / 4 ounces . water = ( 60 / 65 ) * ( 65 / 4 ) = 15 ounces . ans a"
a ) 15 , b ) 30 , c ) 45 , d ) 60 , e ) 90
a
multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 1)
divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)|
other
in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 8 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ?
"this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 8 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . al...
a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60
c
multiply(divide(divide(multiply(7, 8), subtract(11, 7)), 7), 15)
multiply(n1,n2)|subtract(n4,n1)|divide(#0,#1)|divide(#2,n1)|multiply(n0,#3)|
other
the number 523 fbc is divisible by 7 , 89 . then what is the value of f * b * c
lcm of 7 , 8 and 9 is 504 , thus 523 fbc must be divisible by 504 . 523 fbc = 523000 + fbc 523000 divided by 504 gives a remainder of 352 . hence , 352 + fbc = k * 504 . k = 1 fbc = 152 - - > f * b * c = 10 k = 2 fbc = 656 - - > f * b * c = 180 as fbc is three digit number k can not be more than 2 . two answers ? well ...
a ) 504 , b ) 532 , c ) 210 , d ) 180 , e ) 280
d
add(divide(add(divide(523, const_4), divide(523, const_4)), const_2), multiply(add(const_4, const_1), const_10))
add(const_1,const_4)|divide(n0,const_4)|add(#1,#1)|multiply(#0,const_10)|divide(#2,const_2)|add(#4,#3)
general
the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 1.50 + 0.50 ( x βˆ’ 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 2 wheels on each of its other axles ?
"number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 2 = 8 total number of axles = 9 t = 1.50 + 0.50 ( 9 βˆ’ 2 ) = 1.50 + . 5 * 7 = 1.5 + 1.5 = 5 $ answer e"
a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00
e
add(1.50, multiply(0.50, subtract(add(divide(subtract(18, 2), 2), const_1), 2)))
subtract(n3,n2)|divide(#0,n5)|add(#1,const_1)|subtract(#2,n2)|multiply(n1,#3)|add(n0,#4)|
general
indu gave bindu rs . 5000 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ?
"5000 = d ( 100 / 4 ) 2 d = 8 answer : a"
a ) 8 , b ) 2 , c ) 9 , d ) 5 , e ) 1
a
subtract(subtract(multiply(5000, power(add(const_1, divide(4, const_100)), 2)), 5000), multiply(multiply(5000, divide(4, const_100)), 2))
divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)|
gain
a sum of money at simple interest amounts to $ 700 in 3 years and to $ 764 in 4 years . the sum is :
"a $ 508 s . i . for 1 year = $ ( 764 - 700 ) = $ 64 . s . i . for 3 years = $ ( 64 x 3 ) = $ 192 . principal = $ ( 700 - 192 ) = $ 508 ."
a ) $ 508 , b ) $ 698 , c ) $ 398 , d ) $ 549 , e ) $ 675
a
subtract(700, divide(multiply(subtract(764, 700), 3), 4))
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
gain
how many factors does 34 ^ 2 have ?
"36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 5 * 4 = 20 answer c ."
a ) 2 , b ) 8 , c ) 20 , d ) 25 , e ) 26
c
divide(34, multiply(const_10, const_2))
multiply(const_10,const_2)|divide(n0,#0)|
other
how many positive integers between 1 and 100 are there such that they are multiples of 15 ?
"multiples of 15 = 15 , 30,45 , - - - - - 90 number of multiples of 15 = > 90 - 15 / 15 + 1 = 6 answer is e"
a ) 5 , b ) 4 , c ) 7 , d ) 1 , e ) 6
e
divide(subtract(100, 1), 15)
subtract(n1,n0)|divide(#0,n2)|
general
a man walking at a rate of 10 km / hr crosses a bridge in 3 minutes . the length of the bridge is ?
"speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 5 minutes = 50 / 18 * 5 * 60 = 500 m answer is a"
a ) 500 , b ) 650 , c ) 250 , d ) 111 , e ) 236
a
multiply(divide(multiply(10, const_1000), const_60), 3)
multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|
gain
how many paying stones , each measuring 3 * 2 m are required to pave a rectangular court yard 15 m long and 6 m board ?
"15 * 6 = 3 * 2 * x = > x = 15 answer : d"
a ) 99 , b ) 18 , c ) 26 , d ) 15 , e ) 12
d
divide(rectangle_area(15, 6), rectangle_area(3, 2))
rectangle_area(n2,n3)|rectangle_area(n0,n1)|divide(#0,#1)|
physics
on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction w of the total distance did...
if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or...
a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7
e
subtract(divide(lcm(const_2, const_3), 2.8), const_2)
lcm(const_2,const_3)|divide(#0,n1)|subtract(#1,const_2)|
general
what number has a 15 : 1 ratio to the number 10 ?
"15 : 1 = x : 10 x = 10 * 15 x = 150 answer : d"
a ) 130 , b ) 100 , c ) 200 , d ) 150 , e ) 120
d
multiply(10, 15)
multiply(n0,n2)|
other
if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 7 , what is the value of 2 ^ a * 2 ^ b ?
"2 ^ ab = 2 ^ 7 therefore ab = 7 either a = 1 or 7 or b = 7 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 8 = 256 d"
a ) 16 , b ) 32 , c ) 64 , d ) 256 , e ) 128
d
multiply(power(2, const_3.0), 2)
power(n0,n2)|multiply(n0,#0)|
general
a man is 30 years older than his son . in four 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 + 30 . = > 2 ( x + 4 ) = ( x + 30 + 4 ) = > 2 x + 8 = x + 34 = > x = 26 years answer : option e"
a ) 22 years , b ) 23 years , c ) 24 years , d ) 25 years , e ) 26 years
e
divide(subtract(30, 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
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 17 terms of the arithmetic progression ?
"n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 17 . . . we get 17 / 2 [ 2 a + 14 d ] = 17 [ a + 7 d ] = 17 * 10 = 170 . . . answer is d...
a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 270
d
divide(multiply(20, 17), const_2)
multiply(n0,n1)|divide(#0,const_2)|
general
a man invests rs . 4,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ?
"explanation : interest on rs . 4000 at 5 % per annum = ( 4000 Γ— 5 Γ— 1 ) / 100 = rs . 200 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ— 8 Γ— 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 4000 + x ) Γ— 6 Γ— 1 / 100 . = > 200 + 2 x / 25 = ( 4000 + x ) Γ— 6 Γ— ...
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 2000 , e ) none of these
d
multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6)))
multiply(n2,const_1000)|subtract(n3,n1)|subtract(n2,n3)|divide(#1,#2)|multiply(#3,#0)|
gain
a cab driver 5 days income was $ 400 , $ 250 , $ 650 , $ 400 , $ 500 . then his average income is ?
"avg = sum of observations / number of observations avg income = ( 400 + 250 + 650 + 400 + 500 ) / 5 = 440 answer is c"
a ) $ 400 , b ) $ 420 , c ) $ 440 , d ) $ 460 , e ) $ 480
c
divide(add(add(add(add(400, 250), 650), 400), 500), 5)
add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)|
general
calculate the time it will take for a train that is 120 meter long to pass a bridge of 160 meter length , if the speed of the train is 40 km / hour ?
speed = 40 km / hr = 40 * ( 5 / 18 ) m / sec = 11.1111 m / sec total distance = 120 + 160 = 280 meter time = distance / speed = 280 * ( 1 / 11.1111 ) = 25.2 seconds answer : b
a ) 21.2 seconds , b ) 25.2 seconds , c ) 29.2 seconds , d ) 35.2 seconds , e ) 11.2 seconds
b
divide(add(120, 160), divide(40, const_3_6))
add(n0,n1)|divide(n2,const_3_6)|divide(#0,#1)
physics
the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 540 . what is the banker ' s discount ?
"explanation : td = ( bg Γ— 100 ) / tr = ( 540 Γ— 100 ) / ( 6 Γ— 12 ) = ( 90 Γ— 100 ) / 12 = ( 15 Γ— 100 ) 2 / = rs . 750 bg = bd – td = > 540 = bd - 750 = > bd = 540 + 750 = 1290 answer : option d"
a ) 1240 , b ) 1120 , c ) 1190 , d ) 1290 , e ) none of these
d
add(540, divide(multiply(540, const_100), multiply(12, 6)))
multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)|
gain
if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 246 m , what is its area ?
"length = breadth + 23 . therefore , 4 Γ— breadth + 2 Γ— 23 = 246 m β‡’ breadth = 50 m length = 50 + 23 = 73 m area = 73 Γ— 50 = 3650 m 2 answer is b ."
a ) 2510 , b ) 3650 , c ) 2530 , d ) 2515 , e ) 2520
b
rectangle_area(add(divide(subtract(246, multiply(const_2, 23)), const_4), 23), divide(subtract(246, multiply(const_2, 23)), const_4))
multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)|
geometry
a and b can together finish a work in 30 days . they worked together for 20 days and then b left . after another 20 days , a finished the remaining work . in how many days a alone can finish the job ?
"( a + b ) ' s 20 days work = ( 1 / 30 * 20 ) = 2 / 3 . remaining work = ( 1 - 2 / 3 ) = 1 / 3 . now , 1 / 3 work is done by a in 20 days . whole work will be done by a in ( 20 * 3 ) = 60 days . correct option : d"
a ) 40 , b ) 50 , c ) 54 , d ) 60 , e ) none of these
d
divide(multiply(20, 30), subtract(30, 20))
multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)|
physics
how many numbers between 190 and 580 are divisible by 4,5 and 6 ?
every such number must be divisible by l . c . m of 4 , 5,6 i . e , 60 . such numbers are 240,300 , 360,420 , 480,540 . clearly , there are 6 such numbers answer : a
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
a
subtract(floor(divide(580, lcm(lcm(const_4, add(const_4, const_1)), 6))), floor(divide(190, lcm(lcm(const_4, add(const_4, const_1)), 6))))
add(const_1,const_4)|lcm(#0,const_4)|lcm(n3,#1)|divide(n1,#2)|divide(n0,#2)|floor(#3)|floor(#4)|subtract(#5,#6)
general
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 24 sec . what is the length of the platform ?
"speed = 300 / 24 = 25 / 2 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 25 / 2 = > x = 187.5 m . answer : c"
a ) 389 m , b ) 350 m , c ) 187.5 m , d ) 299 m , e ) 219.5 m
c
subtract(multiply(speed(300, 24), 39), 300)
speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)|
physics
what percent is 3 % of 5 % ?
"required percentage = 3 % / 5 % * 100 = 3 / 5 * 100 = 60 % answer is a"
a ) 60 % , b ) 50 % , c ) 75 % , d ) 45 % , e ) 30 %
a
multiply(divide(3, 5), const_100)
divide(n0,n1)|multiply(#0,const_100)|
gain
when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum t of all possible 2 - digit remainders for x / y ?
"ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . t = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b"
a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784
b
divide(59.32, subtract(2, floor(2)))
floor(n1)|subtract(n1,#0)|divide(n0,#1)|
general
perimeter of an equilateral and isosceles is 45 and 40 respectively . at least one of the sides of isosceles is equal to the equilateral . what ' s the base of isosceles triangle ?
10 units as other 2 sides will be 15,15 units . answer : d
['a ) 7', 'b ) 8', 'c ) 9', 'd ) 10', 'e ) 11']
d
subtract(40, multiply(divide(45, const_3), const_2))
divide(n0,const_3)|multiply(#0,const_2)|subtract(n1,#1)
geometry
kamal obtained 76 , 60 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ?
"sol . average = 76 + 60 + 82 + 67 + 85 / 5 ) = ( 375 / 5 ) = 74 . answer c"
a ) 65 , b ) 69 , c ) 74 , d ) 75 , e ) none
c
divide(add(add(add(add(76, 60), 82), 67), 85), add(const_1, const_4))
add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)|
general
a train covers a distance of 12 km in 10 min . if it takes 16 sec to pass a telegraph post , then the length of the train is ?
"speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 16 = 320 m . answer : a"
a ) 320 m , b ) 188 m , c ) 120 m , d ) 178 m , e ) 189 m
a
divide(12, subtract(divide(12, 10), 16))
divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)|
physics
if 12 men and 16 boys can do a piece of work in 5 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy .
"12 m + 16 b - - - - - 5 days 13 m + 24 b - - - - - - - 4 days 60 m + 80 b = 52 m + 96 b 8 m = 16 b = > 1 m = 2 b m : b = 2 : 1 answer : b"
a ) 2 : 9 , b ) 2 : 1 , c ) 2 : 7 , d ) 2 : 5 , e ) 2 : 2
b
divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13)))
multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|
physics
what least value should be replaced by * in 223 * 431 so the number become divisible by 9
"explanation : trick : number is divisible by 9 , if sum of all digits is divisible by 9 , so ( 2 + 2 + 3 + * + 4 + 3 + 1 ) = 15 + * should be divisible by 9 , 15 + 3 will be divisible by 9 , so that least number is 3 . answer : option a"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
a
subtract(431, subtract(431, 431))
subtract(n1,n1)|subtract(n1,#0)|
general
the cost price of a radio is rs . 1340 and it was sold for rs . 1210 , find the loss % ?
"1340 - - - - 130 100 - - - - ? = > 9 % answer : b"
a ) 18 % , b ) 9 % , c ) 22 % , d ) 24 % , e ) 21
b
multiply(divide(subtract(1340, 1210), 1340), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
the probability that a number selected at random from the first 50 natural numbers is a composite number is ?
"the number of exhaustive events = ⁡ ⁰ c ₁ = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : b"
a ) 17 / 65 , b ) 17 / 25 , c ) 17 / 28 , d ) 17 / 65 , e ) 17 / 39
b
divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50)
multiply(const_3,const_6)|subtract(#0,const_1)|multiply(#1,const_2)|divide(#2,n0)|
probability
if x : y = 4 : 7 , find the value of ( 6 x + 2 y ) : ( 5 x – y )
explanation : given : x / y = 4 / 7 ( 6 x + 2 y ) : ( 5 x – y ) = ( 6 * 4 + 2 * 7 ) : ( 5 * 4 – 7 ) 38 : 13 answer : e
a ) 7 : 13 , b ) 38 : 7 , c ) 3 : 13 , d ) 2 : 3 , e ) 38 : 13
e
divide(add(power(6, 2), 2), add(add(4, 7), 2))
add(n0,n1)|power(n2,n3)|add(n3,#1)|add(n3,#0)|divide(#2,#3)
general
ritesh and co . generated revenue of rs . 2,500 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ?
"explanation : given , ritesh and co . generated revenue of rs . 2,500 in 2006 and that this was 12.5 % of the gross revenue . hence , if 2500 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) Γ— 2500 . = > 20,000 . hence , the total revenue by end of 2007 is rs . 10,000 . in 2006 , revenue...
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these
a
multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100)
add(const_1,const_4)|add(const_2,const_4)|divide(n2,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#3,const_100)|add(#6,#5)|multiply(#6,const_2)|add(#8,#4)|divide(#7,#2)|divide(#9,#10)|multiply(#11,const_100)|
gain
according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 15 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ?
"orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 15 oz can = 300 oz / 15 oz = 20 answer d , 20 cans"
a ) 25 , b ) 34 , c ) 50 , d ) 20 , e ) 100
d
divide(multiply(6, 200), divide(200, const_10))
divide(n2,const_10)|multiply(n2,n3)|divide(#1,#0)|
general
the difference between compound interest and simple interest on an amount of $ 15,000 for 2 years is $ 96 . what is the rate of interest per annum ?
"[ 15000 x ( 1 + r / 100 ) ^ 2 - 15000 ] - [ ( 15000 x r x 2 ) / 100 ] = 96 = = > 15000 [ ( 1 + r / 100 ) ^ 2 - 1 - 2 r / 100 ] = 96 = = > 15000 [ ( ( 100 + r ) ^ 2 - 10000 - ( 200 x r ) ) / 10000 ] = 96 = = > r ^ 2 = ( 96 x 2 ) / 3 = 64 = = > r = 8 . rate = 8 % answer d ) 8 %"
a ) 6 % , b ) 7 % , c ) 9 % , d ) 8 % , e ) 5 %
d
divide(96, subtract(power(add(const_1, divide(15,000, const_100)), 2), add(const_1, multiply(2, divide(15,000, const_100)))))
divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5)|
gain
if x + y = 14 , x - y = 60 , for integers of x and y , x = ?
"x + y = 14 x - y = 60 2 x = 74 x = 37 answer is a"
a ) 37 , b ) 35 , c ) 25 , d ) 20 , e ) 42
a
divide(add(14, 60), const_2)
add(n0,n1)|divide(#0,const_2)|
general
if in a race of 100 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by :
"explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 100 * 5 ) / 25 = 20 m hence , a beats b by 20 m . answer a"
a ) 20 m , b ) 16 m , c ) 11 m , d ) 10 m , e ) 15 m
a
multiply(divide(subtract(25, 20), 25), 100)
subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)|
physics
the speed of a car is 90 km in the first hour and 30 km in the second hour . what is the average speed of the car ?
"explanation : s = ( 90 + 30 ) / 2 = 60 kmph b )"
a ) 50 kmph , b ) 60 kmph , c ) 75 kmph , d ) 85 kmph , e ) 90 kmph
b
divide(add(90, 30), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 580 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 800 . how much more do pairs of boots cost than pairs of shoes...
"let x be pair of shoes and y be pair of boots . 22 x + 16 y = 580 . . . eq 1 8 x + 32 y = 800 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1160 8 x = 800 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 800 . . . then we get 32 y = 720 then y = 22.50 differenece between x and y is 12.5...
a ) $ 10.50 , b ) $ 12.50 , c ) $ 11.50 , d ) $ 16.50 , e ) $ 9.50
b
divide(subtract(580, multiply(22, divide(subtract(multiply(580, const_2), 800), subtract(multiply(22, const_2), 8)))), 16)
multiply(n2,const_2)|multiply(n0,const_2)|subtract(#0,n5)|subtract(#1,n3)|divide(#2,#3)|multiply(n0,#4)|subtract(n2,#5)|divide(#6,n1)|
general
the speed of a boat in still water in 65 km / hr and the rate of current is 15 km / hr . the distance travelled downstream in 25 minutes is :
"explanation : speed downstream = ( 65 + 15 ) = 80 kmph time = 25 minutes = 25 / 60 hour = 5 / 12 hour distance travelled = time Γ— speed = ( 2 / 5 ) Γ— 80 = 33.33 km answer : option c"
a ) 55.55 km , b ) 44.44 km , c ) 33.33 km , d ) 22.22 km , e ) 11.11 km
c
multiply(add(65, 15), divide(25, const_60))
add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)|
physics
( √ 27 + √ 243 ) / √ 48 = ?
"( √ 27 + √ 243 ) / √ 48 = ( 3 √ 3 + 9 √ 3 ) / 4 √ 3 = 12 √ 3 / 4 √ 3 = 3 . hence , the correct answer is e ."
a ) 2 √ 2 , b ) 2 √ 3 , c ) 3 √ 2 , d ) 3 √ 3 , e ) 3
e
divide(add(sqrt(27), sqrt(243)), sqrt(48))
sqrt(n0)|sqrt(n1)|sqrt(n2)|add(#0,#1)|divide(#3,#2)|
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 16 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 - - - - - 5 ? - - - - - - 60 12 h rs = 16 + 21 = 37 t = 12 d = 37 * 12 = 444 answer : b"
a ) 2887 , b ) 444 , c ) 877 , d ) 278 , e ) 178
b
add(multiply(divide(60, subtract(21, 16)), 16), multiply(divide(60, subtract(21, 16)), 21))
subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
physics
the volume of a cube is 1728 cc . find its surface ?
a 3 = 1728 = > a = 12 6 a 2 = 6 * 12 * 12 = 864 answer : a
['a ) 864', 'b ) 209', 'c ) 278', 'd ) 276', 'e ) 280']
a
surface_cube(cube_edge_by_volume(1728))
cube_edge_by_volume(n0)|surface_cube(#0)
geometry
you hold some gold in a vault as an investment . over the past year the price of gold increases by 50 % . in order to keep your gold in the vault , you must pay 8 % of the total value of the gold per year . what percentage has the value of your holdings changed by over the past year .
"( 100 % + 50 % ) * ( 100 % - 8 % ) = 150 * 0.92 = 138 % an increase of 38 % your gold holdings have increased in value by 38 % . the answer is d"
a ) 35 % , b ) 36 % , c ) 37 % , d ) 38 % , e ) 40 %
d
subtract(50, divide(50, 8))
divide(n0,n1)|subtract(n0,#0)|
gain
18 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ?
"alcohol in the 18 litres of mix . = 20 % of 18 litres = ( 20 * 18 / 100 ) = 3.6 litres water in it = 18 - 3.6 = 14.4 litres new quantity of mix . = 18 + 3 = 21 litres quantity of alcohol in it = 3.6 litres percentage of alcohol in new mix . = 3.6 * 100 / 21 = 17.14 % answer is b"
a ) 16.67 % , b ) 17.14 % , c ) 18.3 % , d ) 19.75 % , e ) 21.23 %
b
multiply(divide(subtract(add(18, 3), add(multiply(divide(subtract(const_100, 20), const_100), 18), 3)), add(18, 3)), const_100)
add(n0,n2)|subtract(const_100,n1)|divide(#1,const_100)|multiply(n0,#2)|add(n2,#3)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100)|
gain
two isosceles triangles have equal vertical angles and their areas are in the ratio 4 : 9 . find the ratio of their corresponding heights .
we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 4 / 9 - - > height / height = 2 / 3 . answer : e .
['a ) 4 / 5', 'b ) 5 / 4', 'c ) 3 / 2', 'd ) 5 / 7', 'e ) 2 / 3']
e
divide(sqrt(const_4), sqrt(9))
sqrt(const_4)|sqrt(n1)|divide(#0,#1)
geometry
in a certain company , the ratio of male to female employees is 7 : 8 . if 3 more men were hired , this ratio would increase to 8 : 9 . how many male employees are there in the company ?
another approach is to use two variables . let m = present number of males let f = present number of females the ratio of male to female employees is 7 : 8 so , m / f = 7 / 8 cross multiply to get 7 f = 8 m if 3 more men were hired , this ratio would increase to 8 : 9 so , ( m + 3 ) / f = 8 / 9 cross multiply to get 9 ...
a ) 6 , b ) 21 , c ) 27 , d ) 189 , e ) 192
d
divide(multiply(divide(3, subtract(divide(8, 9), divide(7, 8))), 7), 8)
divide(n1,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2)|multiply(n0,#3)|divide(#4,n1)
general
a guy was asked to specify his age in years . he said , β€œ take my age 5 years hence , multiply it by 5 and subtract 5 times of my age 5 years ago and you will know my age . ” what was the age of that guy ?
current age of the guy = a years . then , 5 ( a + 5 ) – 5 ( a – 5 ) = a ( 5 a + 25 ) – ( 5 a – 25 ) = a a = 50 e
a ) 18 , b ) 15 , c ) 33 , d ) 39 , e ) 50
e
add(multiply(5, 5), multiply(5, 5))
multiply(n0,n0)|add(#0,#0)
general
of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 17 boys and 43 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ?
"venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you ...
a ) 5 , b ) 4 , c ) 6 , d ) 8 , e ) 10
a
subtract(subtract(17, subtract(10, 4)), 6)
subtract(n2,n7)|subtract(n4,#0)|subtract(#1,n6)|
other
ram , who is half as efficient as krish , will take 27 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ?
number of days taken by ram to complete task = 27 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 27 days is 27 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work d...
a ) 16 days , b ) 12 days , c ) 9 days , d ) 6 days , e ) 18 days
c
inverse(add(divide(const_1, 27), divide(const_1, divide(27, const_2))))
divide(const_1,n0)|divide(n0,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3)
physics
what is the compound interest on rs : 60,000 for 4 months at the rate of 5 % per annum
"it is monthly compound rate = 5 / 12 % per month 60000 * ( 1 + 5 / 1200 ) ^ 4 - 60000 = 1006.13 answer : c"
a ) 1058.24 , b ) 2006.24 , c ) 1006.13 , d ) 1015.24 , e ) 1014.24
c
divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(4, const_3)))), const_100)
multiply(const_100,const_3)|multiply(const_3,n1)|divide(n1,#1)|multiply(#0,const_100)|multiply(n2,#2)|multiply(#3,#4)|divide(#5,const_100)|
gain
two persons a and b can complete a piece of work in 40 days and 60 days respectively . if they work together , what part of the work will be completed in 6 days ?
"a ' s one day ' s work = 1 / 40 b ' s one day ' s work = 1 / 60 ( a + b ) ' s one day ' s work = 1 / 40 + 1 / 60 = 1 / 24 the part of the work completed in 6 days = 6 ( 1 / 24 ) = 1 / 4 . answer : e"
a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 / 4
e
multiply(6, add(divide(const_1, 40), divide(const_1, 60)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|
physics
at what price must an article costing rs . 95 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ?
"cp = 95 sp = 47.50 * ( 125 / 100 ) = 118.125 mp * ( 95 / 100 ) = 118.125 mp = 125 answer : b"
a ) 130 , b ) 125 , c ) 145 , d ) 135 , e ) 144
b
divide(multiply(add(95, divide(multiply(95, 25), const_100)), const_100), subtract(const_100, 5))
multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)|
gain
a β€œ sophie germain ” prime is any positive prime number p for which 2 p + 1 is also prime . the product of all the possible units digits of sophie germain primes greater than 9 is
"in that case , the sophie prime numbers greater than 9 are 11,23 , 47,59 , . . which yields units digit as 1 , 3,7 and 9 product would be 1 x 3 x 7 x 9 = 189 answer should be d"
a ) 3 , b ) 7 , c ) 21 , d ) 189 , e ) 267
d
multiply(multiply(multiply(subtract(add(add(subtract(9, 1), add(2, 1)), add(2, 1)), const_10), subtract(add(multiply(2, add(subtract(9, 1), add(2, 1))), 1), const_10)), add(2, 1)), subtract(add(multiply(2, add(add(add(subtract(9, 1), add(2, 1)), add(2, 1)), add(2, 1))), 1), multiply(2, const_10)))
add(n0,n1)|multiply(n0,const_10)|subtract(n2,n1)|add(#0,#2)|add(#3,#0)|multiply(n0,#3)|add(n1,#5)|add(#4,#0)|subtract(#4,const_10)|multiply(n0,#7)|subtract(#6,const_10)|add(n1,#9)|multiply(#8,#10)|multiply(#0,#12)|subtract(#11,#1)|multiply(#13,#14)|
general
a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 3 km on foot . his total journey is
"explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 3 = x 5 x + 18 = 6 x x = 18 km answer : option a"
a ) 18 km , b ) 10 km , c ) 12 km , d ) 24 km , e ) 25 km
a
multiply(3, 3)
multiply(n3,n4)|
general
the sum of all consecutive odd integers from βˆ’ 27 to 37 , inclusive , is
"the sum of the odd numbers from - 27 to + 27 is 0 . let ' s add the remaining numbers . 29 + 31 + 33 + 35 + 37 = 5 ( 33 ) = 165 the answer is d ."
a ) 110 , b ) 135 , c ) 150 , d ) 165 , e ) 235
d
add(add(add(add(27, const_2), add(add(27, const_2), const_2)), add(add(add(27, const_2), const_2), const_2)), 37)
add(n0,const_2)|add(#0,const_2)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(n1,#4)|
physics
if 20 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ?
"let 20 % of a = 1 / 3 b then 20 a / 100 = 1 b / 3 a / 5 = b / 3 a / b = 5 / 3 a : b = 5 : 3 answer is c"
a ) 2 : 5 , b ) 1 : 4 , c ) 5 : 3 , d ) 6 : 11 , e ) 2 : 3
c
divide(divide(const_1, const_4), divide(20, const_100))
divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)|
general
a person travels from p to q a speed of 60 km / hr and returns by increasing his speed by 50 % . what is his average speed for both the trips ?
speed on return trip = 150 % of 60 = 90 km / hr . average speed of trip = 60 + 90 / 2 = 150 / 2 = 75 km / hr answer : b
a ) 33 , b ) 75 , c ) 48 , d ) 99 , e ) 21
b
divide(add(multiply(60, add(const_1, divide(50, const_100))), 60), const_2)
divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)
general
( 1.1 + 1.1 + 1.1 + 1.1 ) x 1.1 x 1.1 = ? = ? x 0.121
"explanation : ? = ( 1.1 + 1.1 + 1.1 + 1.1 ) x 1.1 x 1.1 / 0.121 = ( 4.4 x 1.1 x 1.1 ) / 0.121 = 44 answer : option c"
a ) 54 , b ) 33 , c ) 44 , d ) 22 , e ) 26
c
subtract(divide(multiply(1.1, add(1.1, const_1)), const_2), divide(multiply(subtract(1.1, const_1), 1.1), 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
how much is 85 % of 40 is greater than 4 / 5 of 25 ?
"( 85 / 100 ) * 40 – ( 4 / 5 ) * 25 = 14 answer : b"
a ) 22 , b ) 14 , c ) 88 , d ) 12 , e ) 66
b
subtract(multiply(40, divide(85, const_100)), multiply(divide(4, 5), 25))
divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)|
general
( 23341379 x 72 ) = ?
"explanation : 23341379 x 72 = 23341379 ( 70 + 2 ) = ( 23341379 x 70 ) + ( 23341379 x 2 ) = 1633896530 + 46682758 = 1680579288 . answer : option a"
a ) 1680579288 , b ) 1223441288 , c ) 2142579288 , d ) 2142339288 , e ) none of these
a
multiply(23341379, power(add(const_4, const_1), const_4))
add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)|
general
in a fuel station the service costs $ 2.10 per vehicle and every liter of fuel costs $ 0.70 . assuming that you fill up 3 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ?
the service cost of 3 vans and 2 trucks is 5 * 2.10 $ 10.50 the fuel in 3 vans is 3 * 65 = 195 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 481 liters the total fuel cost is 481 * 0.7 = $ 336.70 the total cost is $ 336.70 + $ 10.50 = $ 347.20 the answer is d .
a ) $ 338.50 , b ) $ 341.40 , c ) $ 344.30 , d ) $ 347.20 , e ) $ 350.10
d
add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.7), multiply(multiply(3, 65), 0.7)), multiply(2.1, add(3, 2)))
add(n2,n3)|multiply(n4,n5)|multiply(n2,n4)|divide(#1,const_100)|multiply(n1,#2)|multiply(n0,#0)|add(n4,#3)|multiply(n3,#6)|multiply(n1,#7)|add(#8,#4)|add(#9,#5)
general
following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 15 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ?
price of candy before price increase = 10 / 1.25 = 8 price of soda before price increase = 15 / 1.5 = 10 total price = 8 + 10 = 18 d is the answer
a ) 11 . , b ) 12 . , c ) 13 . , d ) 18 . , e ) 14.5
d
add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(15, const_100), add(const_100, 50)))
add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|multiply(n1,const_100)|divide(#2,#0)|divide(#3,#1)|add(#4,#5)
general
the population of a bacteria colony doubles every day . if it was started 9 days ago with 2 bacteria and each bacteria lives for 11 days , how large is the colony today ?
"9 days ago - 2 8 days ago - 4 7 days ago - 8 6 days ago - 16 5 days ago - 32 4 days ago - 64 3 days ago - 128 2 days ago - 256 yesterday - 512 today - 1024 ans : e"
a ) 512 , b ) 768 , c ) 4096 , d ) 2048 , e ) 1024
e
subtract(power(2, add(9, const_1)), const_1)
add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|
physics
in how many ways can an answer key for a quiz be written if the quiz contains 4 true - false questions followed by 2 multiples - choice questions with 4 answer choices each , if the correct answers to all true - false questions can not be the same ?
"there are 2 ^ 4 = 16 possibilities for the true - false answers . however we need to remove two cases for tttt and ffff . there are 4 * 4 = 16 possibilities for the multiple choice questions . the total number of possibilities is 14 * 16 = 224 . the answer is c ."
a ) 120 , b ) 190 , c ) 224 , d ) 298 , e ) 256
c
multiply(subtract(power(2, 4), 2), multiply(4, 4))
multiply(n2,n2)|power(n1,n0)|subtract(#1,n1)|multiply(#0,#2)|
general
in what time will a train 135 m long cross an electric pole , it its speed be 140 km / hr ?
"speed = 140 * 5 / 18 = 38.8 m / sec time taken = 135 / 38.8 = 3.5 sec . answer : a"
a ) 3.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1
a
divide(135, multiply(140, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
a man buys a cycle for rs . 1200 and sells it at a loss of 10 % . what is the selling price of the cycle ?
"s . p . = 90 % of rs . 1200 = rs . 90 x 1200 / 100 = rs . 1080 answer : option a"
a ) s . 1080 , b ) s . 1140 , c ) s . 1090 , d ) s . 1202 , e ) s . 1092
a
divide(multiply(subtract(const_100, 10), 1200), const_100)
subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|
gain
the average age of 30 students in a class is 14 years . if the age of teacher is also included , the average becomes 15 years , find the age of the teacher .
explanation : if teacher ' s age is 14 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 15 . age of teacher = average age of all + total increase in age = 15 + ( 1 x 30 ) = 45 years answer : c
a ) 43 , b ) 44 , c ) 45 , d ) 47 , e ) 48
c
subtract(add(add(multiply(30, 14), 15), 30), multiply(30, 14))
multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)
general
john and lewis leave city a for city b simultaneously at 6 a . m in the morning driving in two cars at speeds of 40 mph and 60 mph respectively . as soon as lewis reaches city b , he returns back to city a along the same route and meets john on the way back . if the distance between the two cities is 240 miles , how fa...
time taken by lewis to reach city b = 240 / 60 = 4 hours in 4 hours , john travels 40 * 4 = 160 miles so distance at which they meet should be greater than 160 miles . only b satisfies . answer is b .
a ) 150 miles , b ) 160 miles , c ) 140 miles , d ) 170 miles , e ) 200 miles
b
multiply(40, divide(240, 60))
divide(n3,n2)|multiply(n1,#0)
physics
a salesman Γ’ € β„’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4000 . if his remuneration as per new scheme was rs . 700 more than that by the previous schema , his sales were worth ?
[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 700 x = 8000 answer a
a ) s . 8,000 , b ) s . 9,000 , c ) s . 20,000 , d ) s . 10,000 , e ) s . 50,000
a
divide(divide(subtract(subtract(1000, multiply(4000, divide(2.5, const_100))), 700), subtract(divide(5, const_100), divide(2.5, const_100))), 1000)
divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#1,#0)|subtract(n1,#2)|subtract(#4,n4)|divide(#5,#3)|divide(#6,n1)
general
the average weight of 8 persons increases by 3 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ?
total weight increased = ( 8 x 3 ) kg = 24 kg . weight of new person = ( 65 + 24 ) kg = 89 kg . answer : d
a ) 75 kg , b ) 65 kg , c ) 55 kg , d ) 89 kg , e ) 25 kg
d
add(65, multiply(8, 3))
multiply(n0,n1)|add(n2,#0)
general
in a village of 100 households , 85 have at least one dvd player , 90 have at least one cell phone , and 55 have at least one mp 3 player . if x and y are respectively the greatest and lowest possible number of households that have all 3 of these devices , x – y is :
am i missing something here ? ? ? it seems straightforward . . . . . . the obvious maximum that have all 3 is 55 , because you are limited by the smallest number . the minimum is simply the sum of the max of each people who dont have the product , so : 100 - 90 = 10 do n ' t have cell 100 - 85 = 15 do n ' t have dvd an...
a ) 65 , b ) 55 , c ) 45 , d ) 35 , e ) 25
e
subtract(55, subtract(100, add(subtract(100, 55), add(subtract(100, 85), subtract(100, 90)))))
subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|subtract(n3,#5)
general
of the diplomats attending a summit conference , 14 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ?
{ total } = { french } + { russian } - { both } + { neither } { total } = 14 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 180 . answer : e .
a ) 72 , b ) 96 , c ) 108 , d ) 120 , e ) 180
e
divide(subtract(32, 14), subtract(divide(20, const_100), divide(10, const_100)))
divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)
other
in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 80 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the ...
"the percent of tagged fish in the second catch is 2 / 80 * 100 = 2.5 % . we are told that 2.5 % approximates the percent of tagged fish in the pond . since there are 80 tagged fish , then we have 0.025 x = 80 - - > x = 3,200 . answer : d ."
a ) 400 , b ) 625 , c ) 1250 , d ) 3200 , e ) 10 000
d
divide(80, divide(2, 80))
divide(n2,n1)|divide(n0,#0)|
gain
the cost price of an article is 25 % of the marked price . calculate the gain percent after allowing a discount of 50 % .
"sol . let marked price = rs . 100 . then , c . p . = rs . 25 . s . p = rs . 50 . Γ’ Λ† Β΄ gain % = [ 25 / 25 * 100 ] % = 100.0 % . answer c"
a ) 25 % , b ) 50 % , c ) 100 % , d ) 150 % , e ) none
c
multiply(subtract(divide(subtract(const_100, 50), 25), const_1), const_100)
subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)|
gain
the total marks obtained by a student in mathematics and physics is 40 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together .
"let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 40 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 40 + 20 ) / 2 = 30 . answer : e"
a ) 40 , b ) 26 , c ) 27 , d ) 28 , e ) 30
e
divide(add(40, 20), const_2)
add(n0,n1)|divide(#0,const_2)|
general
anup manages to draw 7 circles of equal radii with their centres on the diagonal of a square such that two extreme circles touch two sides of the square and each middle circle touches two circles on either side . find the ratio of radius of the circles to the side of the square .
diagonal of square = ( 2 * r ) * 7 + ( 2 ^ ( 1 / 2 ) * r - r ) * 2 diagonal has 7 squares diameters length + little distance between extreme squares to the end point [ form a square of side r from center of extreme circle to the extreme end to find distance ] distance = 2 ^ ( 1 / 2 ) * r - r d = a * 2 ^ ( 1 / 2 ) a = 2...
['a ) 1 : ( 2 + 6 ^ ( 1 / 2 ) )', 'b ) 1 : ( 4 + 7 ^ ( 1 / 3 ) )', 'c ) ( 2 + 7 ^ ( 1 / 2 ) ) : 1', 'd ) ( 2 + 7 ^ ( 1 / 2 ) ) : 2', 'e ) none of these']
e
divide(sqrt(const_2), multiply(const_2, add(7, sqrt(const_2))))
sqrt(const_2)|add(n0,#0)|multiply(#1,const_2)|divide(#0,#2)
geometry
a light flashes every 20 seconds , how many times will it flash in ? of an hour ?
"1 flash = 20 sec for 1 min = 3 flashes so for 1 hour = 3 * 60 = 180 flashes . answer : d"
a ) 550 , b ) 600 , c ) 650 , d ) 180 , e ) 750
d
divide(const_3600, 20)
divide(const_3600,n0)|
physics
the cost of 3 pens and 5 pencils is rs . 240 . also the cost of one pen and one pencil is in the ratio of 5 : 1 respectively . what is the cost of one dozen pens ?
"explanation : let the cost of one pen is β€˜ 5 x ’ and pencil is β€˜ x ’ 3 x 5 x + 5 x = rs . 240 15 x + 5 x = rs . 240 x = 240 / 20 = 12 : . cost of 1 pen = 5 x = 5 x 12 = 60 : . cost of 12 pens , i . e . ( one dozen ) = 60 x 12 = rs . 720 answer : option b"
a ) rs . 200 , b ) rs . 720 , c ) rs . 300 , d ) rs . 150 , e ) none of these
b
multiply(multiply(3, const_4), divide(240, add(3, 1)))
add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)|
other
the regular price per can of a certain brand of soda is $ 0.55 . if the regular price per can is discounted 25 percent when the soda is purchased in 24 - can cases , what is the price of 70 cans of this brand of soda purchased in 24 - can cases ?
"the discounted price of one can of soda is ( 0.75 ) ( $ 0.55 ) , or $ 0.4125 therefore , the price of 70 cans of soda at the discounted price would be ( 70 ) ( $ 0.4125 ) = 28.875 answer : e"
a ) $ 16.32 , b ) $ 18.00 , c ) $ 21.60 , d ) $ 24.48 , e ) $ 28.87
e
multiply(divide(subtract(const_100, 25), const_100), multiply(0.55, 70))
multiply(n0,n3)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)|
gain
the ratio between the number of sheep and the number of horses at the stewart farm is 2 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ?
"let the number of sheeps and horses be 4 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 2 x = 2 * 8 = 16 . hence b"
a ) 18 , b ) 16 , c ) 32 , d ) 56 , e ) 60
b
multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 2)
multiply(const_2,const_4)|multiply(const_1000,const_12)|multiply(#0,const_10)|multiply(#0,const_100)|add(#2,#3)|add(#4,#1)|divide(#5,n2)|divide(#6,n1)|multiply(n0,#7)|
other
the average monthly salary of 20 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?
"manager ' s monthly salary rs . ( 1600 * 21 - 1500 * 20 ) = rs . 3600 . answer : a"
a ) 3600 , b ) 2988 , c ) 2789 , d ) 2887 , e ) 1297
a
subtract(multiply(add(1500, 100), add(20, const_1)), multiply(1500, 20))
add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
general
a call center has two teams . each member of team a was able to process 6 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ?
"let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 6 calls , so total calls for team a = 30 fraction for team b = 40 / ( 40 + 30 ) = 4 / 7 = answer = c"
a ) 3 / 2 , b ) 3 / 4 , c ) 4 / 7 , d ) 1 / 2 , e ) 1 / 5
c
divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 6)))
multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|divide(#0,#2)|
general
after decreasing 24 % in the price of an article costs rs . 1064 . find the actual cost of an article ?
"cp * ( 76 / 100 ) = 1064 cp = 14 * 100 = > cp = 1400 answer : e"
a ) 1667 , b ) 6789 , c ) 1200 , d ) 6151 , e ) 1400
e
divide(1064, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
a metallic sheet is of rectangular shape with dimensions 48 m x 36 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 :
"explanation clearly , l = ( 48 – 16 ) m = 32 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 32 x 20 x 8 ) m 3 = 5120 m 3 . answer b"
a ) 4830 , b ) 5120 , c ) 6420 , d ) 8960 , e ) none
b
volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8)
multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)|
geometry
300 Γ— ? + ( 12 + 4 ) Γ— 1 / 8 = 602
explanation : = > 300 Γ— ? + ( 12 + 4 ) Γ— 1 / 8 = 602 = > 300 Γ— ? = 602 - ( 12 + 4 ) Γ— 1 / 8 = > 300 Γ— ? = 602 - 2 = 600 = > ? = 600 / 300 = 2 answer : option d
a ) a ) 4 , b ) b ) 3 , c ) c ) 5 , d ) d ) 2 , e ) e ) 8
d
divide(subtract(602, multiply(add(12, 4), divide(1, 8))), 300)
add(n1,n2)|divide(n3,n4)|multiply(#0,#1)|subtract(n5,#2)|divide(#3,n0)
general
if 2 / z = 2 / ( z + 1 ) + 2 / ( z + 36 ) which of these integers could be the value of z ?
"solving for z algebraically in this problem would not be easy . instead , we can follow the hint in the question ( β€œ which of these integers … ” ) and test each answer choice : a . 2 / 0 = 2 / 1 + 2 / 36 incorrect ( division by zero ) b . 2 / 1 = 2 / 2 + 2 / 37 incorrect c . 2 / 2 = 2 / 3 + 2 / 38 incorrect d . 2 / 3 ...
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 6
e
divide(36, add(add(2, 2), 1))
add(n0,n0)|add(n2,#0)|divide(n4,#1)|
general
girl and boy together can complete a piece of work in 35 days while girl alone can complete the same work in 60 days . boy alone will be able to complete the same working in :
subtraction of fraction ab βˆ’ cd = ad βˆ’ cbbd girl and boy finish one work with company = 35 days ( girl + boy ) β€² s one day ’ s work = 135 girl alone finish the same work = 60 days girl β€² s one day ’ s work = 160 b β€² s one day ’ s work = ( a + b ) β€² s one day ’ s work - a β€² s one day ’ s work 135 βˆ’ 160 = 184 hence b alo...
a ) 72 , b ) 75 , c ) 84 , d ) 88 , e ) 90
b
subtract(inverse(subtract(inverse(35), inverse(60))), const_10)
inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)|subtract(#3,const_10)
physics
what is 15 percent of 34
"explanation : it will be 15 % of 34 = ( 15 / 100 ) * 34 = 5.10 answer : option a"
a ) 5.1 , b ) 4.1 , c ) 3.1 , d ) 2.1 , e ) none of these
a
divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 140 . how much was lent at 5 % ?
"( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 140 x = 1000 answer : d"
a ) 2288 , b ) 27669 , c ) 1766 , d ) 1000 , e ) 2871
d
divide(subtract(140, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100)))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,n2)|divide(#2,const_100)|subtract(#0,#1)|subtract(n3,#3)|divide(#5,#4)|
gain
the difference between the local value and the face value of 7 in the numeral 32675149 is ?
"( local value of 7 ) - ( face value of 7 ) = ( 70000 - 7 ) = 69993 d )"
a ) 75142 , b ) 64851 , c ) 5149 , d ) 69993 , e ) none of these
d
subtract(multiply(7, const_1000), 7)
multiply(n0,const_1000)|subtract(#0,n0)|
general
the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 16 , and k < m < r < s < t . if t is 42 , what is the greatest possible value of the median of the 5 integers ?
we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 16 which means that in effect , all numbers are equal to 16 . if the largest number is 42 , it is 26 more than 16 . we need r to be max...
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 22
a
subtract(divide(subtract(multiply(16, 5), 42), const_2), const_2)
multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|subtract(#2,const_2)
general
a rectangular field has a length 10 meters more than it is width . if the area of the field is 231 , what is the length ( in meters ) of the rectangular field ?
"area = l * w = ( l ) * ( l - 10 ) = 231 trial and error : 20 * 10 = 200 ( too low ) 21 * 11 = 231 the length is 21 meters . the answer is a ."
a ) 21 , b ) 23 , c ) 25 , d ) 27 , e ) 29
a
add(10, add(const_0_25, add(const_0_33, divide(divide(231, 10), const_2))))
divide(n1,n0)|divide(#0,const_2)|add(#1,const_0_33)|add(#2,const_0_25)|add(n0,#3)|
geometry
a cycle is bought for rs . 600 and sold for rs . 1080 , find the gain percent ?
"600 - - - - 180 100 - - - - ? = > 80 % answer : d"
a ) 22 , b ) 20 , c ) 90 , d ) 80 , e ) 11
d
multiply(divide(subtract(1080, 600), 600), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
how many integers are between 3 and 86 / 7 , inclusive ?
"86 / 7 = 12 . xx we are not concerned about the exact value of 86 / 7 as we just need the integers . since the values are small , we can write down the integers . the different integers between 3 and 86 / 7 would be 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 total number of integers = 10 option c if we need to find the ...
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
c
add(subtract(divide(86, 7), 3), const_1)
divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)|
general
a man goes downstream at 15 kmph , and upstream 8 kmph . the speed of the stream is
"speed of the stream = 1 / 2 ( 15 - 8 ) kmph = 3.5 kmph . correct option : b"
a ) 0 kmph , b ) 3.5 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph
b
divide(subtract(15, 8), const_2)
subtract(n0,n1)|divide(#0,const_2)|
physics
if 12 people contributed a total of $ 20.00 toward a gift and each of them contributed at least $ 1.00 , then the maximum possible amount any one person could have contributed is
d for me 11 people with 1 $ each - > maximum = 9
a ) $ 1.00 , b ) $ 1.25 , c ) $ 5.00 , d ) $ 9.00 , e ) $ 20.00
d
subtract(20, multiply(subtract(12, const_1), 1))
subtract(n0,const_1)|multiply(n2,#0)|subtract(n1,#1)
general
the diagonals of a rhombus are 12 cm and 20 cm . find its area ?
"1 / 2 * 12 * 20 = 120 answer : c"
a ) 329 , b ) 288 , c ) 120 , d ) 238 , e ) 31
c
rhombus_area(12, 20)
rhombus_area(n0,n1)|
geometry
the majority owner of a business received 25 % of the profit , with each of 4 partners receiving 25 % of the remaining profit . if the majority owner and two of the owners combined to receive $ 76,875 , how much profit did the business make ?
"let p be the total profit . p / 4 + 1 / 2 * ( 3 p / 4 ) = p / 4 + 3 p / 8 = 5 p / 8 = $ 76,875 p = $ 123,000 the answer is e ."
a ) $ 98,000 , b ) $ 106,000 , c ) $ 112,000 , d ) $ 118,000 , e ) $ 123,000
e
divide(multiply(const_100, multiply(const_100, add(const_1, 4))), add(divide(25, const_100), multiply(multiply(divide(25, const_100), subtract(const_1, divide(25, const_100))), const_2)))
add(const_1,n1)|divide(n0,const_100)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,const_100)|multiply(#1,#3)|multiply(#5,const_2)|add(#1,#6)|divide(#4,#7)|
gain