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on dividing a number by 5 , we get 2 as quotient and 0 as remainder . on dividing the same number by 4 , what will be the remainder ?
"number = 5 * 2 + 0 = 10 4 ) 10 ( 2 8 - - - - - - - - 2 required number = 2 . answer : c"
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
c
reminder(multiply(5, 2), 4)
multiply(n0,n1)|reminder(#0,n3)|
general
two tains of equal lengths take 10 seconds and 16 seconds respectively to cross a telegraph post . if the length of each train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?
"sol . speed of the first train = [ 120 / 10 ] m / sec = 12 m / sec . speed of the second train = [ 120 / 16 ] m / sec = 7.5 m / sec . relative speed = ( 12 + 7.5 ) = m / sec = 19.5 m / sec . ∴ required time = ( 120 + 120 ) / 19.5 secc = 12.3 sec . answer d"
a ) 12 , b ) 14 , c ) 16 , d ) 12.3 , e ) 18
d
divide(add(120, 120), add(divide(120, 16), divide(120, 10)))
add(n2,n2)|divide(n2,n1)|divide(n2,n0)|add(#1,#2)|divide(#0,#3)|
physics
how many paying stones , each measuring 2 m * 1 m are required to pave a rectangular court yard 30 m long and 16 m board ?
30 * 16 = 2 * 1 * x = > x = 240 answer : a
['a ) 240', 'b ) 18', 'c ) 26', 'd ) 17', 'e ) 12']
a
divide(rectangle_area(30, 16), rectangle_area(2, 1))
rectangle_area(n2,n3)|rectangle_area(n0,n1)|divide(#0,#1)
physics
the circumferences of two circles are 268 meters and 380 meters . find the difference between the areas of the larger and the smaller circles .
"let the radii of the smaller and the larger circles be s m and l m respectively . 2 ∏ s = 268 and 2 ∏ l = 380 s = 268 / 2 ∏ and l = 380 / 2 ∏ difference between the areas = ∏ l 2 - ∏ s 2 = ∏ { 1902 / ∏ 2 - 1342 / ∏ 2 } = 1902 / ∏ - 1342 / ∏ = ( 190 - 134 ) ( 190 + 134 ) / ∏ = ( 56 ) ( 324 ) / ( 22 / 7 ) = ( 18144 ) / ...
a ) 388.15 , b ) 2992 , c ) 4312.5 , d ) 2887.27 , e ) 5773.09
e
subtract(circle_area(divide(380, multiply(const_2, const_pi))), circle_area(divide(268, multiply(const_2, const_pi))))
multiply(const_2,const_pi)|divide(n1,#0)|divide(n0,#0)|circle_area(#1)|circle_area(#2)|subtract(#3,#4)|
geometry
if teena is driving at 55 miles per hour and is currently 7.5 miles behind roe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of roe ?
"this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . roe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena ca...
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105
d
multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60)
subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)|
physics
the difference between the local value and the face value of 3 in the numeral 62975143 is
"explanation : ( local value of 3 ) - ( face value of 3 ) = ( 3 - 3 ) = 0 d )"
a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 5
d
subtract(multiply(3, const_1000), 3)
multiply(n0,const_1000)|subtract(#0,n0)|
general
the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 36 ?
first , let ' s translate the expression 21 # 36 , using the definition given : 21 # 36 = 21 Γ— 24 Γ— 27 Γ— 30 Γ— 33 Γ— 36 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 36 = 3 ^ 6 ( 7 Γ— 8 Γ— 9 Γ— 10 Γ— 11 Γ— 12 ) now let ' s replace each consecutive integer with its prime facto...
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20
c
add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2)))
add(const_2,const_2)|add(n0,#0)|add(#1,#0)|add(#2,#1)
general
a grocer has a sale of rs . 435 , rs . 927 , rs . 855 , rs . 230 and rs . 562 for 5 consecutive days . how much sale must he have in the sixth day so that he gets an average sale of rs . 625 ?
"explanation : total sale for 5 days = rs . ( 435 + 927 + 855 + 230 + 562 ) = rs . 3009 . required sale = rs . [ ( 625 x 6 ) - 3009 ] = rs . ( 3750 - 3009 ) = rs . 741 . answer : a"
a ) rs 741 , b ) rs 5991 , c ) rs 6001 , d ) rs 6991 , e ) none of these
a
subtract(multiply(add(5, const_1), 625), add(add(add(add(435, 927), 855), 230), 562))
add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|
general
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 13 cm .
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 13 ) = 247 cm 2 answer : d"
a ) 335 cm 2 , b ) 885 cm 2 , c ) 285 cm 2 , d ) 247 cm 2 , e ) 725 cm 2
d
quadrilateral_area(13, 18, 20)
quadrilateral_area(n2,n1,n0)|
physics
the monthly rent of a shop of dimension 10 feet Γ— 8 feet is rs . 2400 . what is the annual rent per square foot of the shop ?
"sol . monthly rent per square feet = 2400 / ( 10 * 8 ) = 30 & annual rent per square feet = 12 * 30 = 360 answer : d"
a ) 43 , b ) 56 , c ) 68 , d ) 360 , e ) 92
d
multiply(add(const_10, const_2), divide(2400, rectangle_area(10, 8)))
add(const_10,const_2)|rectangle_area(n0,n1)|divide(n2,#1)|multiply(#0,#2)|
geometry
find compound interest on $ 12000 at 15 % per annum for 2 years 4 months , compounded annually .
"time = 2 years 4 months = 2 ( 4 / 12 ) years = 2 ( 1 / 3 ) years . amount = $ [ 12000 x ( 1 + Β­ ( 15 / 100 ) ) 2 x ( 1 + ( ( 1 / 3 ) * 15 ) / 100 ) ] = $ [ 12000 * ( 23 / 20 ) * ( 23 / 20 ) * ( 21 / 20 ) ] = $ 16663.50 . : . c . i . = rs . ( 16663.50 - 12000 ) = $ 4663.50 answer a ."
a ) 4663.5 , b ) 4203.5 , c ) 4303.5 , d ) 4403.5 , e ) 4103.5
a
add(add(multiply(12000, divide(15, const_100)), multiply(add(12000, multiply(12000, divide(15, const_100))), divide(15, const_100))), multiply(add(add(12000, multiply(12000, divide(15, const_100))), multiply(add(12000, multiply(12000, divide(15, const_100))), divide(15, const_100))), divide(divide(15, const_100), const...
divide(n1,const_100)|divide(#0,const_3)|multiply(n0,#0)|add(n0,#2)|multiply(#3,#0)|add(#2,#4)|add(#3,#4)|multiply(#6,#1)|add(#5,#7)|
gain
after decreasing 24 % in the price of an article costs rs . 320 . find the actual cost of an article ?
"cp * ( 76 / 100 ) = 320 cp = 4.21 * 100 = > cp = 421 answer : e"
a ) 400 , b ) 520 , c ) 651 , d ) 525 , e ) 421
e
divide(320, subtract(const_1, divide(24, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)|
gain
in a group of 800 people , 1 / 5 play at least one instrument , 32 play two or more . what is the probability that one student play exactly one instrument ?
"p ( playing 2 or more instruments ) = 32 / 800 = 1 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 1 / 5 - 1 / 25 = 4 / 25 . answer b ."
a ) 2 / 125 , b ) 4 / 25 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 5
b
divide(subtract(divide(multiply(800, 1), 5), 32), 800)
multiply(n0,n1)|divide(#0,n2)|subtract(#1,n3)|divide(#2,n0)|
general
teacher took exam for english , average for the entire class was 80 marks . if we say that 10 % of the students scored 95 marks and 20 % scored 90 marks then calcualte average marks of the remaining students of the class .
lets assume that total number of students in class is 100 and required average be x . then from the given statement we can calculate : ( 10 * 95 ) + ( 20 * 90 ) + ( 70 * x ) = ( 100 * 80 ) 70 x = 8000 - ( 950 + 1800 ) = 5250 answer : d x = 75 .
a ) 60 , b ) 70 , c ) 65 , d ) 75 , e ) 80
d
divide(subtract(subtract(80, multiply(divide(10, const_100), 95)), multiply(divide(20, const_100), 90)), subtract(const_1, add(divide(10, const_100), divide(20, const_100))))
divide(n1,const_100)|divide(n3,const_100)|add(#0,#1)|multiply(n2,#0)|multiply(n4,#1)|subtract(n0,#3)|subtract(const_1,#2)|subtract(#5,#4)|divide(#7,#6)
general
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 2000 more than d , what is b ' s share ?
"let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 2000 = > x = 2000 . b ' s share = rs . 2 x = 2 * 2000 = rs . 4000 . answer : a"
a ) a ) 4000 , b ) b ) 2900 , c ) c ) 2000 , d ) d ) 2393 , e ) e ) 2009
a
multiply(multiply(subtract(4, 3), 2000), 3)
subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)|
general
what is the square root of 324 ?
"a square root of a number is a number times itself equals the original number . ex ) the square root of 81 is 9 because 9 times itself ( 9 ) equals the original number ( 81 ) . 18 * 18 = 324 324 divided by x = 18 x = 18 ( a ) 18"
a ) a ) 18 , b ) b ) 9 , c ) c ) 45 , d ) d ) 62 , e ) e ) 81
a
circle_area(divide(324, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
other
a and b together have rs . 1210 . if of a ' s amount is equal to of b ' s amount , how much amount does b have ?
4 / 15 a = 2 / 5 b a = ( 2 / 5 x 15 / 4 ) b a = 3 / 2 b a / b = 3 / 2 a : b = 3 : 2 . therefore , b ' s share = rs . ( 1210 x 2 / 5 ) = rs . 484 . answer is a .
a ) 484 , b ) 460 , c ) 550 , d ) 664 , e ) none of them
a
multiply(divide(const_2, add(const_3, const_2)), 1210)
add(const_2,const_3)|divide(const_2,#0)|multiply(n0,#1)
general
4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 10 weavers in 10 days ?
"1 weaver can weave 1 mat in 4 days . 10 weavers can weave 10 mats in 4 days . 10 weavers can weave 25 mats in 10 days . the answer is d ."
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30
d
multiply(10, multiply(10, divide(4, multiply(4, 4))))
multiply(n0,n0)|divide(n0,#0)|multiply(n3,#1)|multiply(n3,#2)|
gain
if a trader sold two cars each at rs . 404415 and gains 15 % on the first and loses 15 % on the second , then his profit or loss percent on the whole is ?
"explanation : sp of each car is rs . 404415 , he gains 15 % on first car and losses 15 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 15 ) ( 15 ) / 100 % = 2.25 % answer is d"
a ) 1.44 % , b ) 2.02 % , c ) 1.04 % , d ) 2.25 % , e ) 3.40 %
d
divide(multiply(15, 15), const_100)
multiply(n1,n1)|divide(#0,const_100)|
gain
robert ate 10 chocolates , nickel ate 5 chocolates . how many more chocolates did robert ate than nickel ?
10 - 5 = 5 . answer is d
a ) a ) 4 , b ) b ) 7 , c ) c ) 9 , d ) d ) 5 , e ) e ) 2
d
subtract(10, 5)
subtract(n0,n1)|
general
if a + b = βˆ’ 9 , and a = 30 / b , what is the value of a ^ 2 + b ^ 2 ?
"a ^ 2 + b ^ 2 should make you think of these formulas : ( a + b ) ( a + b ) = a ^ 2 + b ^ 2 + 2 ab we already know ( a + b ) = - 9 and a * b = 30 ( a + b ) ( a + b ) = ( - 9 ) ( - 9 ) = a ^ 2 + b ^ 2 + 2 * ( 30 ) a ^ 2 + b ^ 2 = 81 - 60 = 21 answer : b"
a ) 31 , b ) 21 , c ) 41 , d ) 61 , e ) 51
b
subtract(power(9, 2), multiply(2, 30))
multiply(n1,n2)|power(n0,n2)|subtract(#1,#0)|
general
a basket contains 6 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ?
"the total number of ways to choose 2 apples is 6 c 2 = 15 the number of ways that include the spoiled apple is 5 c 1 = 5 p ( the spoiled apple is included ) = 5 / 15 = 1 / 3 the answer is a ."
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6
a
divide(choose(subtract(6, 1), 1), choose(6, 2))
choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)|
probability
if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 60 , what is the value of f ( 5 ) ?
"f ( x ) = 2 x ^ 2 + y f ( 2 ) = 60 = > 2 * ( 2 ) ^ 2 + y = 60 = > 8 + y = 60 = > y = 52 f ( 5 ) = 2 * ( 5 ) ^ 2 + 52 = 102 answer e"
a ) 104 , b ) 60 , c ) 52 , d ) 50 , e ) 102
e
add(subtract(60, multiply(power(2, 2), 2)), multiply(2, power(5, 2)))
power(n4,n0)|power(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|subtract(n3,#3)|add(#2,#4)|
general
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 884 in 4 years . the sum is :
"s . i . for 1 year = rs . ( 884 - 815 ) = rs . 69 . s . i . for 3 years = rs . ( 69 x 3 ) = rs . 207 . principal = rs . ( 815 - 207 ) = rs . 608 . answer : option a"
a ) s . 608 , b ) s . 690 , c ) s . 698 , d ) s . 700 , e ) s . 760
a
subtract(815, divide(multiply(subtract(884, 815), 3), 4))
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
gain
if n divided by 7 has a remainder of 2 , what is the remainder when 2 times n is divided by 7 ?
"as per question = > n = 7 p + 2 for some integer p hence 2 n = > 14 q + 4 = > remainder = > 4 for some integer q alternatively = > n = 2 > 2 n = > 4 = > 4 divided by 7 will leave a remainder 4 hence d"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6
d
multiply(2, 2)
multiply(n1,n2)|
general
sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6900 ?
"total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6900 x 6 ) - 34009 ] = rs . ( 41400 - 34009 ) = rs . 7391 answer : c"
a ) 4966 , b ) 6477 , c ) 7391 , d ) 2676 , e ) 1881
c
multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6900), 5)
add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5)|
general
if 60 percent of 600 is 50 percent of x , then x = ?
"0.6 * 600 = 0.5 * x x = 6 / 5 * 600 = 720"
a ) 600 , b ) 720 , c ) 820 , d ) 800 , e ) 920
b
divide(multiply(60, 600), 50)
multiply(n0,n1)|divide(#0,n2)|
general
the length of the bridge , which a train 145 m long and traveling at 45 km / hr can cross in 30 sec is ?
"speed = 45 * 5 / 18 = 25 / 2 m / sec . time = 30 sec let the length of bridge be x meters . then , ( 145 + x ) / 30 = 25 / 2 x = 230 m . answer : option a"
a ) 230 , b ) 240 , c ) 245 , d ) 250 , e ) 255
a
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 145)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
a $ 89.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ?
"listed selling price of chair = 89.95 $ discounted selling price of chair = 59.95 $ discount = 89.95 - 59.95 = 30 $ % decrease in price of chair = ( 30 / 89.95 ) * 100 % = 33 % approx answer e"
a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 33 %
e
multiply(divide(subtract(89.95, 59.95), 89.95), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
general
a money lender lent rs . 1000 at 3 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 390 . find the number of years .
( 1000 xtx 3 / 100 ) + ( 1400 xtx 5 / 100 ) = 390 Γ’ † ’ t = 3.9 answer d
a ) 3.5 , b ) 3.75 , c ) 4 , d ) 3.9 , e ) 4.5
d
divide(390, add(divide(multiply(3, 1000), const_100), divide(multiply(1400, 5), const_100)))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|divide(n4,#4)
gain
company c produces toy trucks at a cost of $ 5.00 each for the first 100 trucks and $ 3.50 for each additional truck . if 500 toy trucks were produced by company c and sold for $ 12.00 each , what was company c ’ s gross profit ?
"cost of 500 trucks : ( 100 * 5 ) + ( 400 * 3.5 ) = 500 + 1400 = $ 1900 revenue : 500 * 12 = $ 6000 profit : 6000 - 1900 = $ 4100 option c is correct"
a ) $ 2,250 , b ) $ 2,500 , c ) $ 4,100 , d ) $ 3,250 , e ) $ 4,500
c
floor(divide(subtract(subtract(multiply(500, 12.00), multiply(5.00, 100)), multiply(subtract(500, 100), 3.50)), const_1000))
multiply(n3,n4)|multiply(n0,n1)|subtract(n3,n1)|multiply(n2,#2)|subtract(#0,#1)|subtract(#4,#3)|divide(#5,const_1000)|floor(#6)|
gain
a certain rectangular window is three times as long as it is wide . if its perimeter is 32 feet , what are its dimensions in terms of length by width ?
2 x + 2 y = 32 x + y = 16 x + 3 x = 16 4 x = 16 x = 4 answer d
['a ) 12 by 2', 'b ) 11 by 3', 'c ) 10.5 by 3.5', 'd ) 12 by 4', 'e ) 9 by 3']
d
rectangle_area(32, const_4)
rectangle_area(n0,const_4)
geometry
a car averages 50 mph for the first 4 hours of a trip and averages 80 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ?
let the time for which car averages 80 mph = t 65 * ( t + 4 ) = 50 * 4 + 80 t = > 15 t = 60 = > t = 4 total duration of the trip = 4 + 4 = 8 answer a
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 4
a
add(divide(subtract(multiply(65, const_4), multiply(50, 4)), subtract(80, 65)), 4)
multiply(n3,const_4)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)
general
a sum of rs . 1000 at simple interest amounts to rs . 1192 in 4 years . the s . i . for 1 year is :
s . i . for 4 years = rs . ( 1192 - 1000 ) = rs . 192 . s . i . for 1 year = rs . 192 / 4 = rs . 48 . answer : option e
a ) rs . 45 , b ) rs . 70 , c ) rs . 39 , d ) rs . 72 , e ) rs . 48
e
multiply(divide(subtract(divide(1192, const_10), const_100), 4), const_10)
divide(n1,const_10)|subtract(#0,const_100)|divide(#1,n2)|multiply(#2,const_10)
gain
a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 900 grams . find the profit percent .
"explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 900 100 + g / 100 + 0 = 1000 / 900 100 + g = 10 / 9 * 100 g = 11.11 answer : b"
a ) 12.11 , b ) 11.11 , c ) 13.11 , d ) 14.11 , e ) 15.11
b
multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 900), 900), const_100)
add(const_1,const_4)|add(#0,#0)|multiply(#1,const_100)|subtract(#2,n0)|divide(#3,n0)|multiply(#4,const_100)|
gain
if 0.5 % of a = 75 paise , then the value of a is ?
"answer ∡ 0.5 / 100 of a = 75 / 100 ∴ a = rs . ( 75 / 0.5 ) = rs . 150 correct option : a"
a ) rs . 150 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none
a
divide(75, 0.5)
divide(n1,n0)|
gain
if | 4 x + 6 | = 46 , what is the sum of all the possible values of x ?
"there will be two cases 4 x + 6 = 46 or 4 x + 6 = - 46 = > x = 10 or x = - 12 sum of both the values will be - 12 + 10 = - 2 answer is b"
a ) 2 , b ) - 2 , c ) 4 , d ) - 5 , e ) 6
b
divide(subtract(46, 6), 4)
subtract(n2,n1)|divide(#0,n0)|
general
( √ 27 + √ 192 ) / √ 54 = ?
"( √ 27 + √ 192 ) / √ 54 = ( 3 √ 3 + 8 √ 3 ) / 3 √ 3 * 2 = 11 √ 3 / 3 √ 3 * 2 = 11 / ( 3 * √ 2 ) = 11 / ( 3 * √ 2 ) . hence , the correct answer is d ."
a ) 2 √ 2 , b ) 2 √ 3 , c ) 3 √ 2 , d ) 11 / ( 3 * √ 2 ) , e ) √ 2
d
divide(add(sqrt(27), sqrt(192)), sqrt(54))
sqrt(n0)|sqrt(n1)|sqrt(n2)|add(#0,#1)|divide(#3,#2)|
general
60 % of x is greater than 1 / 3 rd of x by 110 . what is x ?
"6 x / 10 - x / 3 = 110 4 x / 15 = 110 x = 412.5 answer : e"
a ) 100.5 , b ) 200.5 , c ) 300.5 , d ) 350.5 , e ) 412.5
e
divide(110, subtract(divide(60, const_100), divide(1, 3)))
divide(n0,const_100)|divide(n1,n2)|subtract(#0,#1)|divide(n3,#2)|
general
a train traveling at 72 kmph crosses a platform in 32 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ?
"speed of train = 72 * ( 5 / 18 ) = 20 m / s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 32 = 640 m . so , length of platform = 640 - 3...
a ) 280 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined
a
subtract(multiply(divide(multiply(72, const_1000), const_3600), 32), multiply(divide(multiply(72, const_1000), const_3600), 18))
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)|
physics
a man buys a cycle for rs . 1400 and sells it at a loss of 5 % . what is the selling price of the cycle ?
"since , c . p = 1400 loss % = ( c . p - s . p ) / c . p * 100 5 = ( 1400 - s . p ) / 1400 * 100 so , after solving answer = 1330 . answer : e"
a ) s . 1090 , b ) s . 1160 , c ) s . 1190 , d ) s . 1202 , e ) s . 1330
e
divide(multiply(subtract(const_100, 5), 1400), const_100)
subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)|
gain
the population of a bacteria colony doubles every day . if it was started 5 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ?
"2 ^ 5 ( 2 ) = 2 ^ 6 = 64 the answer is c"
a ) 512 , b ) 768 , c ) 64 , d ) 2048 , e ) 4096
c
subtract(power(2, add(5, const_1)), const_1)
add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)|
physics
a goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 36 seconds . what is the length of the goods train ?
"explanation : speed = [ 72 x ( 5 / 18 ) ] m / sec = 20 m / sec . time = 36 sec . let the length of the train be x metres . then , [ ( x + 250 ) / 36 ] = 20 = > x + 250 = 720 = > x = 470 . answer : a"
a ) 470 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) none of these
a
multiply(subtract(36, divide(250, multiply(const_0_2778, 72))), multiply(const_0_2778, 72))
multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2)|
physics
if an integer n is to be selected at random from 1 to 900 , inclusive , what is probability n ( n + 1 ) will be divisible by 9 ?
for n ( n + 1 ) to be a multiple of 9 , either n or n + 1 has to be a multiple of 9 . thus n must be of the form 9 k or 9 k - 1 . the probability is 2 / 9 . the answer is d .
a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 9 , e ) 2 / 3
d
divide(const_2, 9)
divide(const_2,n3)
general
find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 100 of money ?
"p = 100 ( 100 / 10 ) 2 = > p = 2000 answer : e"
a ) 1500 , b ) 6000 , c ) 2500 , d ) 1400 , e ) 2000
e
multiply(multiply(divide(100, multiply(10, 2)), const_100), multiply(10, 2))
multiply(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|multiply(#2,#0)|
general
45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 45 meters deep by working for 6 hours ?
"45 workers * 8 hours / 30 meters = x * 6 / 45 x = 90 total workers 90 - 45 = 45 new workers the answer is c ."
a ) 25 , b ) 35 , c ) 45 , d ) 55 , e ) 65
c
subtract(multiply(multiply(45, divide(8, 6)), divide(45, 30)), 45)
divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0)|
physics
if p / q = 4 / 5 , then the value of 4 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ?
"answer given exp . = 4 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 4 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 4 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 4 / 7 + 6 / 14 = 4 / 7 + 3 / 7 = 7 / 7 = 1 . correct option : c"
a ) 3 / 7 , b ) 34 , c ) 1 , d ) 2 , e ) 3
c
add(divide(4, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5))))
divide(n2,n3)|divide(n0,n1)|add(n4,#1)|subtract(n4,#1)|divide(#3,#2)|add(#0,#4)|
general
patrick purchased 70 pencils and sold them at a loss equal to the selling price of 20 pencils . the cost of 70 pencils is how many times the selling price of 70 pencils ?
"say the cost price of 70 pencils was $ 70 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 70 - 70 p = 20 p - - > p = 7 / 9 . ( cost price ) / ( selling price ) = 1 / ( 7 / 9 ) = 9 / 7 = 1.28 . answer : d ."
a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.28 , e ) 1.35
d
inverse(divide(70, add(70, 20)))
add(n0,n1)|divide(n0,#0)|inverse(#1)|
general
3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 4 men and 5 women will do that task .
3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 4 m + 5 w = 13 w answer is 13 / 14 answer : a
a ) 13 / 14 , b ) 13 / 10 , c ) 13 / 18 , d ) 13 / 16 , e ) 13 / 11
a
divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 4), 5), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2))
subtract(n1,n3)|subtract(n2,n0)|divide(#0,#1)|multiply(n4,#2)|multiply(n2,#2)|add(n5,#3)|add(n3,#4)|divide(#5,#6)
physics
40 % of x is greater than 1 / 3 rd of x by 110 . what is x ?
"4 x / 10 - x / 3 = 110 1 x / 15 = 110 x = 1650 answer : d"
a ) 1100 , b ) 2100 , c ) 1300 , d ) 1650 , e ) 1400
d
divide(110, subtract(divide(40, const_100), divide(1, 3)))
divide(n0,const_100)|divide(n1,n2)|subtract(#0,#1)|divide(n3,#2)|
general
the ratio of boarders to day students at a school was originally 5 to 12 . however , after a number of new boarders join the initial 330 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ?
let x be the number of new boarders . the ratio changed from 5 : 12 up to 1 : 2 = 6 : 12 . 330 / ( 330 + x ) = 5 / 6 x = 66 the answer is b .
a ) 48 , b ) 66 , c ) 72 , d ) 80 , e ) 84
b
subtract(divide(multiply(divide(330, 5), 12), 2), 330)
divide(n2,n0)|multiply(n1,#0)|divide(#1,n4)|subtract(#2,n2)
other
a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 70 meters , what is the height of the building ( in meters ) ?
"the height : length ratio will be equal in both cases . 18 / 45 = x / 70 x = 28 the answer is c ."
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36
c
multiply(70, divide(18, 45))
divide(n0,n1)|multiply(n2,#0)|
physics
the average weight of a group of boys is 20 kg . after a boy of weight 33 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ?
"let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 33 kg joins the group , total weight of boys = 20 x + 33 so 20 x + 33 = 21 ( x + 1 ) = > x = 12 . answer : a"
a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10
a
add(subtract(33, add(20, 1)), 1)
add(n0,n2)|subtract(n1,#0)|add(#1,n2)|
general
presently the ratio between the ages of dan and james is 6 : 5 . after 4 years dan will be 28 . what is the present age of james ?
let the present ages dan and james be 6 x years and 5 x years respectively 6 x + 4 = 28 6 x = 24 x = 4 kim ' s age = 5 x = 20 years answer is a
a ) 20 , b ) 19 , c ) 21 , d ) 18 , e ) 22
a
subtract(28, multiply(const_4, const_2))
multiply(const_2,const_4)|subtract(n3,#0)
other
the sum of two numbers is 62 , and one of them is 12 more than the other . what are the two numbers ?
"in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 62 : word problem = 62 the line over x + 12 is a grouping symbol called a vinculum . it saves...
a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 42 , e ) 25 - 37
e
divide(subtract(62, 12), const_2)
subtract(n0,n1)|divide(#0,const_2)|
general
a tank is filled by 3 pipes a , b , c in 5 hours . pipe c is twice as fast as b and b is twice as fast as a . how much will pipe a alone take to fill the tank ?
"suppose pipe a alone take x hours to fill the tank then pipe b and c will take x / 2 and x / 4 hours respectively to fill the tank . 1 / x + 2 / x + 4 / x = 1 / 5 7 / x = 1 / 5 x = 35 hours answer is b"
a ) 25 hr , b ) 35 hr , c ) 40 hr , d ) 20 hr , e ) 50 hr
b
multiply(add(add(multiply(const_2, const_2), const_2), const_1), 5)
multiply(const_2,const_2)|add(#0,const_2)|add(#1,const_1)|multiply(n1,#2)|
physics
if 12 men and 16 boys can do a piece of work in 5 days ; 13 men and 24 boys can do it in 4 days , then the ratio of the daily work done by a man to that of a boy is ?
"let 1 man ' s 1 day work = x and 1 boy ' s 1 day work = y . then , 12 x + 16 y = 1 / 5 and 13 x + 24 y = 1 / 4 solving these two equations , we get : x = 1 / 100 and y = 1 / 200 required ratio = x : y = 1 / 100 : 1 / 200 = 2 : 1 . answer : a"
a ) 2 : 1 , b ) 2 : 8 , c ) 2 : 9 , d ) 2 : 5 , e ) 2 : 2
a
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
a certain company ’ s profit in 1996 was 20 percent greater than its profit in 1995 , and its profit in 1997 was 20 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ?
"profit in 1995 - 100 profit in 1996 - 120 % increae profit in 1997 in comparison to 1995 = 20 + 120 * 20 % = 44 correct option : d"
a ) 5 % , b ) 18 % , c ) 33 % , d ) 44 % , e ) 38 %
d
multiply(subtract(multiply(add(divide(20, const_100), const_1), add(const_1, divide(20, const_100))), const_1), const_100)
divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
a cycle is bought for rs . 670 and sold for rs . 880 , find the gain percent ?
"explanation : 670 - - - - 210 100 - - - - ? = > 31 % answer : a"
a ) 31 % , b ) 35 % , c ) 54 % , d ) 38 % , e ) 80 %
a
multiply(divide(subtract(880, 670), 670), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
a certain no . when divided by 20 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
"explanation : 20 + 25 = 45 / 15 = 3 ( remainder ) a"
a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 9
a
subtract(subtract(subtract(20, 25), const_4), const_2)
subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)|
general
the s . i . on a certain sum of money for 6 years at 14 % per annum is half the c . i . on rs . 7000 for 2 years at 7 % per annum . the sum placed on s . i . is ?
c . i . = [ 7000 * ( 1 + 7 / 100 ) 2 - 7000 ] = ( 7000 * 11 / 10 * 11 / 10 - 7000 ) = rs . 1014.3 . sum = ( 507.15 * 100 ) / ( 6 * 14 ) = rs . 603.75 answer : a
a ) 603.75 , b ) 555.75 , c ) 569.55 , d ) 256.25 , e ) 563.23
a
divide(divide(subtract(multiply(7000, power(add(const_1, divide(7, const_100)), 2)), 7000), 2), multiply(6, divide(14, const_100)))
divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)|
gain
selling an kite for rs . 30 , a shop keeper gains 20 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ?
explanation : marked price = rs . 30 c . p . = 100 / 120 * 30 = rs . 25 sale price = 90 % of rs . 30 = rs . 27 required gain % = 0.2 / 25 * 100 = 8 % . answer : a
a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 20 %
a
multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(20, const_100))), divide(multiply(30, const_100), add(20, const_100))), const_100)
add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100)
gain
a fruit seller sells mangoes at the rate of rs . 10 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 %
"explanation : 85 : 10 = 105 : x x = ( 10 Γ— 105 / 85 ) = rs 12.35 option d"
a ) rs 8.81 , b ) rs 9.35 , c ) rs 10.35 , d ) rs 12.35 , e ) none of these
d
divide(multiply(10, add(const_100, 5)), subtract(const_100, 15))
add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)|
gain
a certain galaxy is known to comprise approximately 3 x 10 ^ 11 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ?
"total no . of stars on galaxy = 3 * 10 ^ 11 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 3 * 10 ^ 11 / 50 * 10 ^ 6 = 30 * 10 ^ 3 / 5 = 6000 therefore answer is b"
a ) 800 , b ) 6,000 , c ) 8,000 , d ) 12,000 , e ) 80,000
b
multiply(divide(multiply(divide(multiply(3, 10), 50), power(10, const_4)), const_1000), 3)
multiply(n0,n1)|power(n1,const_4)|divide(#0,n3)|multiply(#2,#1)|divide(#3,const_1000)|multiply(n0,#4)|
general
a hostel had provisions for 250 men for 32 days . if 50 men left the hostel , how long will the food last at the same rate ?
"a hostel had provisions for 250 men for 32 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 32 250 Γ— 32 = 200 x 5 Γ— 32 = 4...
a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 65
b
divide(multiply(250, 32), subtract(250, 50))
multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)|
gain
a cube has a volume of 216 cubic feet . if a similar cube is twice as long , twice as wide , and twice as high , then the volume , in cubic feet of such cube is ?
"volume = 216 = side ^ 3 i . e . side of cube = 6 new cube has dimensions 12 , 12 , and 12 as all sides are twice of teh side of first cube volume = 12 * 12 * 12 = 1728 square feet answer : option b"
a ) 24 , b ) 1728 , c ) 64 , d ) 80 , e ) 100
b
volume_cube(multiply(const_2, cube_edge_by_volume(216)))
cube_edge_by_volume(n0)|multiply(#0,const_2)|volume_cube(#1)|
geometry
how many different values of positive integer x , for which | x + 5 | < x , are there ?
"answer e i opted to put the random value option . i used 0 , 5 , - 5 and the the extreme of 40 and - 40 . . i was able to solve it in 1 : 09 e"
a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16
e
add(5, 5)
add(n0,n0)|
general
if a speaks the truth 80 % of the times , b speaks the truth 60 % of the times . what is the probability that they tell the truth at the same time
explanation : probability that a speaks truth is 80 / 100 = 0.8 probability that b speaks truth is 60 / 100 = 0.6 since both a and b are independent of each other so probability of a intersection b is p ( a ) Γ— p ( b ) = 0.8 Γ— 0.6 = 0.48 answer : b
a ) 0.49 , b ) 0.48 , c ) 0.41 , d ) 0.482 , e ) 0.411
b
multiply(divide(80, multiply(multiply(const_4, const_5), const_5)), divide(60, multiply(multiply(const_4, const_5), const_5)))
multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)
gain
tickets numbered from 1 to 27 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ?
"here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 ] p ( e ) = n ( e ) / n ( s ) = 9 / 27 = 1 / 3 the answer is a ."
a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7
a
divide(divide(27, 3), 27)
divide(n1,n2)|divide(#0,n1)|
general
a man can row downstream at the rate of 12 km / hr and upstream at 6 km / hr . find man ' s rate in still water and the rate of current ?
"explanation : rate of still water = 1 / 2 ( 12 + 6 ) = 9 km / hr rate of current = 1 / 2 ( 12 - 6 ) = 3 km / hr answer : option c"
a ) 9,6 , b ) 6,3 , c ) 9,3 , d ) 6,6 , e ) none of these
c
divide(subtract(12, 6), const_2)
subtract(n0,n1)|divide(#0,const_2)|
gain
if a is an integer greater than 7 but less than 25 and b is an integer greater than 34 but less than 44 , what is the range of a / b ?
"the way to approach this problem is 7 < a < 25 and 34 < b < 41 minimum possible value of a is 8 and maximum is 24 minimum possible value of b is 35 and maximum is 40 range = max a / min b - min a / max b ( highest - lowest ) 24 / 35 - 8 / 40 = 17 / 35 hence d"
a ) 21 / 34 , b ) 1 / 12 , c ) 15 / 64 , d ) 17 / 35 , e ) 7 / 6
d
subtract(divide(subtract(25, const_1), add(34, const_1)), divide(add(7, const_1), subtract(44, const_1)))
add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)|
general
if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 55 , what is the compound interest on the same sum at the rate and for the same time ?
"explanation : sum = ( 55 * 100 ) / ( 2 * 5 ) = rs . 550 amount = [ 550 * ( 1 + 5 / 100 ) 2 ] = rs . 606.375 c . i . = ( 606.375 - 550 ) = rs . 56.37 answer : a"
a ) s . 56.37 , b ) s . 53.22 , c ) s . 56.219 , d ) s . 51.18 , e ) s . 51.11
a
subtract(add(add(divide(multiply(divide(55, multiply(divide(5, const_100), 2)), 5), const_100), divide(55, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(55, multiply(divide(5, const_100), 2)), 5), const_100), divide(55, multiply(divide(5, const_100), 2))), 5), const_100)), divide(55, m...
divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|
gain
if 2 ^ ( 2 w ) = 8 ^ ( w βˆ’ 4 ) , what is the value of w ?
2 ^ ( 2 w ) = 8 ^ ( w βˆ’ 4 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w βˆ’ 4 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 12 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 12 w = 12 the answer is d .
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15
d
divide(multiply(const_12, log(2)), log(2))
log(n0)|multiply(#0,const_12)|divide(#1,#0)
general
a is twice as good as b . and together they finish a piece of work in 18 days . in how many days will a alone finish the work
( a ' s 1 day work ) : ( b ' s 1 day work ) = 2 : 1 a + b 1 day work = 1 / 18 a ' s 1 day work = ( 1 / 18 ) * ( 2 / 3 ) = 1 / 27 a alone can finish the work in 27 days answer is b
a ) 25 , b ) 27 , c ) 15 , d ) 18 , e ) 21
b
inverse(multiply(divide(inverse(18), add(const_2, const_1)), const_2))
add(const_1,const_2)|inverse(n0)|divide(#1,#0)|multiply(#2,const_2)|inverse(#3)
physics
what is the local value of 7 in the difference of 100889 and local value of 3 in 28943712 ?
"100889 - ( local value of 3 ) = 100889 - 3000 = 97889 local value of 7 in 97889 = 7000 a"
a ) 7000 , b ) 700 , c ) 70000 , d ) 70 , e ) 7
a
reminder(multiply(reminder(subtract(28943712, reminder(28943712, const_1000)), multiply(const_10, const_1000)), 100889), multiply(const_10, const_1000))
multiply(const_10,const_1000)|reminder(n3,const_1000)|subtract(n3,#1)|reminder(#2,#0)|multiply(n1,#3)|reminder(#4,#0)|
general
everyone shakes hands with everyone else in a room . total number of handshakes is 55 . number of persons = ?
"in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 55 or n ( n - 1 ) = 110 or n = 11 answer is ( c )"
a ) 14 , b ) 12 , c ) 11 , d ) 15 , e ) 16
c
divide(divide(multiply(55, const_2), const_3), const_4)
multiply(n0,const_2)|divide(#0,const_3)|divide(#1,const_4)|
general
what will be the fraction of 4 %
"explanation : 4 * 1 / 100 = 1 / 25 . friends i know it is quite simple , but trust me while solving percentage questions in hurry we use to do these types of mistake only . so i recommend you to have a bit practise of this option d"
a ) 1 / 20 , b ) 1 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these
d
divide(circle_area(divide(4, const_2)), const_2)
divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)|
gain
the least number which must be subtracted from 670 to make it exactly divisible by 9 is :
"on dividing 670 by 9 , we get remainder = 4 therefore , required number to be subtracted = 4 answer : c"
a ) a ) 2 , b ) b ) 3 , c ) c ) 4 , d ) d ) 5 , e ) e ) 6
c
subtract(670, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9))
add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)|
general
three unbiased coins are tossed . what is the probability of getting 2 heads and 2 tail ?
"let , h - - > head , t - - > tail here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting 3 heads then e = { thh , hth , hht , tht , tth } p ( e ) = n ( e ) / n ( s ) = 5 / 8 answer is a"
a ) 5 / 8 , b ) 1 / 4 , c ) 3 / 8 , d ) 7 / 8 , e ) 1 / 8
a
negate_prob(divide(const_1, power(const_2, const_3)))
power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)|
probability
in an election a candidate who gets 84 % of the votes is elected by a majority of 476 votes . what is the total number of votes polled ?
"let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 84 ) % of x = 16 % of x 84 % of x - 16 % of x = 476 68 x / 100 = 476 x = 476 * 100 / 68 = 700 answer is b"
a ) 630 , b ) 700 , c ) 535 , d ) 450 , e ) 815
b
divide(476, divide(subtract(84, subtract(const_100, 84)), const_100))
subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|
gain
3 numbers are in the ratio 5 : 6 : 8 . the sum of its longest and smallest numbers equals the sum of the third number and 49 . find the third number ?
let the numbers be 5 x , 6 x , 8 x . largest number = 8 x . smallest number = 5 x . third number = 6 x . 8 x + 5 x = 6 x + 49 7 x = 49 = > x = 7 6 x = 42 = > third number is 42 . answer : d
a ) a ) 54 , b ) b ) 75 , c ) c ) 48 , d ) d ) 42 , e ) e ) 63
d
multiply(divide(49, subtract(add(5, 8), 6)), 6)
add(n1,n3)|subtract(#0,n2)|divide(n4,#1)|multiply(n2,#2)
general
the price of an item changed from $ 120 to $ 100 . then later the price decreased again from $ 100 to $ 80 . which of the two decreases was larger in percentage term ?
first decrease in percent part / whole = ( 120 - 100 ) / 120 = 0.17 = 17 % second decrease in percent part / whole = ( 100 - 80 ) / 100 = 0.20 = 20 % the second decrease was larger in percent term . the part were the same in both cases but the whole was smaller in the second decrease . answer b
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %
b
subtract(const_100, multiply(divide(80, 100), const_100))
divide(n3,n1)|multiply(#0,const_100)|subtract(const_100,#1)
gain
what is the next number : 3 , 11 , 83 , __
"3 ^ 0 + 2 = 3 3 ^ 2 + 2 = 11 3 ^ 4 + 2 = 83 3 ^ 6 + 2 = 731 the answer is b ."
a ) 631 , b ) 731 , c ) 831 , d ) 849 , e ) 901
b
subtract(subtract(subtract(multiply(83, const_10), const_100), 3), 3)
multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)|
general
a man sells a car to his friend at 14 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was :
"explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 Γ£ β€” 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 14 % . now s . p = rs . 45000 and loss = 14 % c . p . rs . [ 100 / 86 Γ£ β€” 45000 ] = rs . 52325.58 . correct option : c"
a ) rs . 22325.58 , b ) rs . 32325.58 , c ) rs . 52325.58 , d ) rs . 62325.58 , e ) none of these
c
divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 14))
add(n2,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|divide(#1,#0)|multiply(#3,const_100)|divide(#4,#2)|
gain
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 960 gm . for a kg . his gain is … % .
"his percentage gain is 100 * 40 / 960 as he is gaining 40 units for his purchase of 960 units . so 4.16 % . answer : a"
a ) 4.16 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 %
a
multiply(subtract(inverse(divide(960, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100)
add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
if two sides of a triangle have lengths 3 and 8 , which of the following could be the perimeter of the triangle ? i . 9 ii . 15 iii . 19
the length of any side of a triangle must be larger than the positive difference of the other two sides , but smaller than the sum of the other two sides . therefore : ( 8 - 3 ) < { third side } < ( 8 + 3 ) . 5 < { third side } < 11 . perimeter = { third side } + 8 + 3 = { third side } + 11 . 16 < { perimeter } < 22 . ...
['a ) none', 'b ) i only', 'c ) ii only', 'd ) iii only', 'e ) i , ii , and iii']
d
triangle_perimeter(8, 8, 3)
triangle_perimeter(n0,n1,n1)
geometry
increasing the original price of an article by 10 percent and then increasing the new price by 10 percent is equivalent to increasing the original price by
"1.1 * 1.1 * x = 1.21 * x the answer is d ."
a ) 20 % , b ) 20.5 % , c ) 20.8 % , d ) 21 % , e ) 21.8 %
d
multiply(const_100, subtract(multiply(add(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), const_1))
divide(n0,const_100)|add(#0,const_1)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|
gain
a can run 1.5 km distance in 2 min 20 seconds , while b can run this distance in 2 min 30 sec . by how much distance can a beat b ?
"a takes time 2.20 minutes = 140 sec b takes time 2.30 minutes = 150 sec diffrence = 150 - 140 = 10 sec now we are to find distance covered in 10 sec by b 150 sec = 1500 m 1 sec = 10 m 10 sec = 10 x 10 = 100 m answer : b"
a ) 90 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) 190 m
b
subtract(multiply(1.5, const_1000), multiply(divide(multiply(1.5, const_1000), add(multiply(2, const_60), 30)), add(multiply(1.5, const_60), 20)))
multiply(n0,const_1000)|multiply(n0,const_60)|multiply(n3,const_60)|add(n2,#1)|add(n4,#2)|divide(#0,#4)|multiply(#3,#5)|subtract(#0,#6)|
physics
a 6 % stock yields 8 % . the market value of the stock is ?
"for an income of rs . 8 , investment = rs . 100 . for an income of rs 6 , investment = rs . \ inline \ left ( \ frac { 100 } { 8 } \ times 6 \ right ) = rs . 75 market value of rs . 100 stock = rs . 75 . answer : b"
a ) rs . 11 , b ) rs . 75 , c ) rs . 17 , d ) rs . 16 , e ) rs . 22
b
multiply(divide(const_100, 8), 6)
divide(const_100,n1)|multiply(n0,#0)|
gain
if y is the smallest positive integer such that 6300 multiplied by y is the square of an integer , then y must be
"6300 = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 * 7 to be perfect square , we need to multiply by at least 7 . the answer is b ."
a ) 5 , b ) 7 , c ) 15 , d ) 35 , e ) 45
b
multiply(add(const_2, const_3), const_2)
add(const_2,const_3)|multiply(#0,const_2)|
geometry
if the farmer sells 75 of his chickens , his stock of feed will last for 20 more days than planned , but if he buys 100 more chickens , he will run out of feed 15 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ?
let v denote the volume of feed one chicken consumes per day . then the total volume of feed in stock will be v βˆ— d βˆ— c where d is the number of days the feed will last if the number of chickens does not change and cc is the current number of chickens . from the question it follows that v ( d + 20 ) ( c βˆ’ 75 ) = vdc v ...
a ) 60 , b ) 120 , c ) 240 , d ) 275 , e ) 300
e
multiply(divide(multiply(20, divide(75, divide(subtract(100, 75), subtract(20, 15)))), subtract(20, divide(75, divide(subtract(100, 75), subtract(20, 15))))), divide(subtract(100, 75), subtract(20, 15)))
subtract(n2,n0)|subtract(n1,n3)|divide(#0,#1)|divide(n0,#2)|multiply(n1,#3)|subtract(n1,#3)|divide(#4,#5)|multiply(#6,#2)
general
if the average of t , b , c , 14 and 15 is 12 . what is the average value of t , b , c and 29
t + b + c + 14 + 15 = 12 * 5 = 60 = > t + b + c = 60 - 29 = 31 t + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
d
divide(multiply(12, add(const_2, const_3)), const_4)
add(const_2,const_3)|multiply(n2,#0)|divide(#1,const_4)
general
a square of 2 unit is given and four circles are made on its corners having equal radius of each . and another circle of equal radius is made on the center of the square . find out the area of square which is not covered by circles ?
let the radius of circle is ' r ' area of the circle which is at corner = ( 22 / 7 * r ^ 2 ) / 4 hence the area of all four circles are = 22 / 7 * r ^ 2 areaof circle which is made on center of suare is = 22 / 7 * r ^ 2 total area of all circles are = 2 * 22 / 7 * r ^ 2 . . . . . . . . . . . . . . . . . . . . . . . . ....
['a ) 0.76', 'b ) 0.86', 'c ) 0.96', 'd ) 1.16', 'e ) 1.26']
b
subtract(square_area(2), multiply(circle_area(divide(divide(2, const_2), const_2)), const_4))
divide(n0,const_2)|square_area(n0)|divide(#0,const_2)|circle_area(#2)|multiply(#3,const_4)|subtract(#1,#4)
geometry
sachin is younger than rahul by 6 years . if the ratio of their ages is 7 : 9 , find the age of sachin
"if rahul age is x , then sachin age is x - 6 , so ( x - 6 ) / x = 7 / 9 = > 9 x - 42 = 7 x = > 2 x = 42 = > x = 21 so sachin age is 21 - 6 = 15 answer : c"
a ) 24.58 , b ) 14 , c ) 15 , d ) 24.9 , e ) 24.1
c
multiply(divide(6, subtract(9, 7)), 7)
subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)|
other
31 of the scientists that attended a certain workshop were wolf prize laureates , and 13 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientist...
"lets solve by creating equation . . w = 31 . . total = 52 . . not w = 52 - 31 = 21 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 21 or x = 9 . . so who won nobel but not wolf = x + 3 = 12 . . but people who won both w and n = 13 . . so total who won n = 12 + 13 = 25 ....
a ) a ) 11 , b ) b ) 18 , c ) c ) 24 , d ) d ) 25 , e ) d ) 36
d
add(add(3, divide(subtract(subtract(51, 31), 3), const_2)), 13)
subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3)|
physics
a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area covered by the two roads ?
area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads = 1000 + 600 - 25 = 1575 answer : option c
['a ) 400', 'b ) 1550', 'c ) 1575', 'd ) 1600', 'e ) 1625']
c
add(rectangle_area(200, 5), rectangle_area(120, 5))
rectangle_area(n0,n2)|rectangle_area(n1,n2)|add(#0,#1)
geometry
if p # q denotes the least common multiple of p and q , then q = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ?
"there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , ...
a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72
c
add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24)
multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2)|add(n3,#3)|
general
the difference between a two - digit number and the number obtained by interchanging the digits is 9 . what is the difference between the two digits of the number ?
"suppose the two - digit number be 10 x + y . then we have been given l 0 x + y – ( 10 y + x ) = 9 β‡’ 9 x – 9 y = 9 β‡’ x – y = 1 hence , the required difference = 1 note that if the difference between a two - digit number and the number obtained by interchanging the digits is d , then the difference between the two digit...
a ) 8 , b ) 2 , c ) 7 , d ) can not be determined , e ) none of these
e
divide(9, subtract(const_10, const_1))
subtract(const_10,const_1)|divide(n0,#0)|
general
81 , 64 , 27 , 16 . . .
"81 / 3 = 27 64 / 4 = 16 27 / 3 = 9 answer : c"
a ) 12 , b ) 4 , c ) 9 , d ) 3 , e ) 1
c
subtract(negate(16), multiply(subtract(64, 27), divide(subtract(64, 27), subtract(81, 64))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
set x consists of all two - digit primes and set y consists of all positive odd multiples of 7 less than 100 . if the two sets are combined into one , what will be the range of the new set ?
set x = { 11 , 13 , 17 , . . . . . . . . . . . . . , 83 , 89 , 97 } set y = { 7 , 21 , 35 , . . . . . . . . . . . . . . . , 77 , 91 , } combining two sets , say set z set z = { 7 , 11 , 13 , 17,21 , . . . . . . . . . . . . . . . . . . . , 77 , 83 , 89 , 91 , 97 } range = max value - min value range ( z ) = 97 - 7 = 90 ...
a ) 84 , b ) 89 , c ) 90 , d ) 92 , e ) 95
c
subtract(subtract(100, const_3), 7)
subtract(n1,const_3)|subtract(#0,n0)
general