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a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 7.5 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the med...
"the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ...
a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 23 / 32 , e ) 3 / 4
d
divide(add(const_1, divide(subtract(multiply(divide(7.5, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2)
divide(n2,n3)|divide(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,const_1)|subtract(#3,const_1)|divide(const_1,#5)|divide(#4,#6)|add(#7,const_1)|divide(#8,const_2)|
general
barbata invests $ 1400 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ?
let the additional invested amount for 8 % interest be x ; equation will be ; 1400 + 0.05 * 2400 + x + 0.08 x = 1400 + x + 0.06 ( 2400 + x ) 0.05 * 1400 + 0.08 x = 0.06 x + 0.06 * 1400 0.02 x = 1400 ( 0.06 - 0.05 ) x = 1400 * 0.01 / 0.02 = 700 ans : a
a ) 700 , b ) 300 , c ) 1000 , d ) 360 , e ) 2400
a
divide(subtract(multiply(divide(6, const_100), 1400), multiply(1400, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100)))
divide(n3,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)
general
a person has 12004 coins of silver . how much would he need to add so that he can distribute equally among his 7 children in whole numbers ?
he would need to add one coin more so 12005 is fully divisible by 7 and each would get 1715 coins . so the correct answer is a
a ) 1 coin , b ) 2 coins , c ) 3 coins , d ) 4 coins , e ) coins
a
subtract(multiply(add(floor(divide(12004, 7)), const_1), 7), 12004)
divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)
general
a cistern 7 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is :
"area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 7 x 1.25 ) + 7 x 4 ] m 2 = 55.5 m 2 . answer : option d"
a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 55.5 m 2 , e ) 57 m 2
d
add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 7))), multiply(4, 7))
divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)|
physics
a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 162 . find the unknown rate of two blankets ?
"10 * 162 = 1620 3 * 100 + 5 * 150 = 1050 1620 – 1050 = 570 answer : d"
a ) 420 , b ) 550 , c ) 490 , d ) 570 , e ) 457
d
subtract(multiply(const_10, 162), add(multiply(3, 100), multiply(5, 150)))
multiply(n4,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)|
general
a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 12 days , a finished the remaining work . in how many days a alone can finish the job ?
"a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 12 days whole work will be done by a in 12 * 4 / 3 = 16 days answer is d"
a ) 10 , b ) 25 , c ) 60 , d ) 16 , e ) 20
d
divide(multiply(12, 40), subtract(40, 10))
multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)|
physics
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 20 % , then the cost price of the article is :
"explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 120 = rs . 466.7 answer : c"
a ) 500 , b ) 277 , c ) 466.7 , d ) 297 , e ) 111
c
divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 20))
add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)|
gain
the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 4 . the sum of numbers is :
"let the numbers be 1 x and 4 x . then , their l . c . m = 4 x . so , 4 x = 48 or x = 12 . the numbers are 12 and 48 . hence , required sum = ( 12 + 48 ) = 60 . answer : e"
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 60
e
divide(multiply(1, 48), 4)
multiply(n0,n1)|divide(#0,n2)|
other
the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half yearly is
"solution amount of rs . 100 for 1 year when compounded half yearly = rs . [ 100 x ( 1 + 3 / 100 ) 2 ] = rs . 106.09 % . effective rate = ( 106.09 - 100 ) % = rs . 6.09 % . answer d"
a ) 6.06 % , b ) 6.07 % , c ) 6.08 % , d ) 6.09 % , e ) none
d
add(add(divide(6, const_2), divide(6, const_2)), divide(multiply(divide(6, const_2), divide(6, const_2)), const_100))
divide(n0,const_2)|add(#0,#0)|multiply(#0,#0)|divide(#2,const_100)|add(#1,#3)|
gain
on a certain day , tim invested $ 500 at 10 percent annual interest , compounded annually , and lana invested 1,000 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ ...
"compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 500 ( 1.10 ) ^...
a ) $ 2.5 , b ) $ 15 , c ) $ 50 , d ) $ 100 , e ) $ 105
a
subtract(subtract(multiply(500, power(add(const_1, divide(10, const_100)), 2)), 500), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 1,000), 1,000))
divide(n1,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|power(#2,n4)|power(#3,n4)|multiply(n0,#4)|multiply(n2,#5)|subtract(#6,n0)|subtract(#7,n2)|subtract(#8,#9)|
gain
two trains of equal are running on parallel lines in the same direction at 49 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ?
"let the length of each train be x m . then , distance covered = 2 x m . relative speed = 49 - 36 = 13 km / hr . = 13 * 5 / 18 = 65 / 18 m / sec . 2 x / 36 = 65 / 18 = > x = 65 . answer : c"
a ) 50 , b ) 88 , c ) 65 , d ) 55 , e ) 22
c
divide(multiply(36, divide(multiply(subtract(49, 36), const_1000), const_3600)), const_2)
subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)|
general
evaluate : 1222343 - 12 * 3 * 2 = ?
"according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 1222343 - 12 * 3 * 2 = 122343 - 8 = 122336 correct answer a"
a ) 122336 , b ) 145456 , c ) 122347 , d ) 126666 , e ) 383838
a
subtract(1222343, multiply(multiply(12, 3), 2))
multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|
general
the ratio of boarders to day students at a school was originally 2 to 5 . however , after a number of new boarders join the initial 60 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 2 : 5 = 4 : 10 up to 1 : 2 = 5 : 10 . 60 / ( 60 + x ) = 4 / 5 x = 15 the answer is b ."
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30
b
subtract(divide(multiply(divide(60, 2), 5), 2), 60)
divide(n2,n0)|multiply(n1,#0)|divide(#1,n4)|subtract(#2,n2)|
other
a candidate got 33 % of the votes polled and he lost to his rival by 833 votes . how many votes were cast ?
"let x be the total number of votes . 0.33 x + 833 = 0.67 x 0.34 x = 833 x = 833 / 0.34 = 2450 the answer is b ."
a ) 2250 , b ) 2450 , c ) 2650 , d ) 2850 , e ) 3050
b
divide(833, subtract(subtract(const_1, divide(33, const_100)), divide(33, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|
gain
calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 18 % and its breadth is decreased by 16 % ?
"let l and b be 100 each 100 * 100 = 10000 l increase by 18 % = 118 b decrease by 16 % = 84 118 * 84 = 9912 0.88 % decrease answer : b"
a ) 5.88 % decrease , b ) 0.88 % decrease , c ) 0.88 % increase , d ) 1.88 % decrease , e ) 2.88 % decrease
b
multiply(divide(subtract(multiply(add(const_100, 18), subtract(const_100, 16)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100)
add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,#1)|multiply(#5,const_100)|
geometry
a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 6 km in stationary water ?
"speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 2 + 6 ) = 4 kmhr time taken in stationary = 6 / 4 = 1 hrs 30 min answer : d"
a ) 40 minutes , b ) 1 hour , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min
d
divide(6, divide(add(multiply(divide(1, 10), const_60), divide(2, 2)), const_2))
divide(n0,n0)|divide(n2,n3)|multiply(#1,const_60)|add(#0,#2)|divide(#3,const_2)|divide(n4,#4)|
physics
jane makes toy bears . when she works with an assistant , she makes 71 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ?
"let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 68.4 bears per 36 hours a week or 1.9 bears / hr ( [ 40 bears * 1.71 ] / [ 40 hrs * . 90 ] ) . [ ( 1.9 - 1 ) / 1 ] * 100 % = 90 % answer : c"
a ) 20 % , b ) 80 % , c ) 90 % , d ) 180 % , e ) 200 %
c
multiply(divide(10, subtract(subtract(const_100, 71), 10)), const_100)
subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)|
physics
34 . the side surface of a cylinder can is rolled with a rectangular plate . if the height of a cylinder tank is 8 feet and the perimeter of the circular base is 6 feet , what is the diagonal of the rectangular plate ?
think of a can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both , use pythagoreans theorem or properties of 3 - 4 - 5 triangles to solve for the hypo...
['a ) 10', 'b ) 12', 'c ) 8', 'd ) 14', 'e ) 9']
a
sqrt(add(power(8, const_2), power(6, const_2)))
power(n1,const_2)|power(n2,const_2)|add(#0,#1)|sqrt(#2)
geometry
a train 120 meters long completely crosses a 480 meters long bridge in 55 seconds . what is the speed of the train is ?
s = ( 120 + 480 ) / 45 = 600 / 55 * 18 / 5 = 39 answer : c
a ) 32 , b ) 545 , c ) 39 , d ) 40 , e ) 09
c
divide(divide(add(120, 480), const_1000), divide(55, const_3600))
add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)
physics
the length of the bridge , which a train 170 metres long and travelling at 45 km / hr can cross in 30 seconds , is :
"speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 170 + x ) / 30 = 25 / 2 = > 2 ( 170 + x ) = 750 = > x = 205 m . answer : option a"
a ) 205 , b ) 210 , c ) 215 , d ) 250 , e ) 260
a
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 170)
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
physics
employees of a certain company are each to receive a unique 6 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , and 5 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ?
"there are 4 ! ways to make codes starting with 12 . there are 4 ! ways to make codes starting with 24 . the number of codes is 2 * 4 ! = 48 . the answer is a ."
a ) 48 , b ) 56 , c ) 64 , d ) 72 , e ) 80
a
multiply(multiply(2, 3), 2)
multiply(n3,n4)|multiply(n3,#0)|
general
the diagonals of a rhombus are 22 cm and 30 cm . find its area ?
"1 / 2 * 22 * 30 = 330 answer : d"
a ) 358 , b ) 329 , c ) 350 , d ) 330 , e ) 317
d
rhombus_area(22, 30)
rhombus_area(n0,n1)|
geometry
in an examination , a pupil ’ s average marks were 63 per paper . if he had obtained 20 more marks for his geography paper and 2 more marks for his history paper , his average per paper would have been 65 . how many papers were there in the examination ?
solution let the number pf papers be x . then , 63 x + 20 + 2 = 65 x 65 x - 63 x = 22 2 x = 22 x = 11 . answer d
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
d
divide(add(20, 2), subtract(65, 63))
add(n1,n2)|subtract(n3,n0)|divide(#0,#1)
general
if 15 % of 40 is greater than 25 % of a number by 2 , then find the number is ?
"15 / 100 * 40 - 25 / 100 * x = 2 or x / 4 = 4 so x = 16 answer c"
a ) 20 , b ) 15 , c ) 16 , d ) 25 , e ) 30
c
divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100))
divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)|
gain
what is the rate percent when the simple interest on rs . 800 amount to rs . 144 in 4 years ?
"144 = ( 800 * 4 * r ) / 100 r = 4.5 % answer : d"
a ) 5 % , b ) 8 % , c ) 3 % , d ) 4.5 % , e ) 1 %
d
divide(multiply(const_100, 144), multiply(800, 4))
multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|
gain
if p / q = 2 / 7 , then 2 p + q = ?
"let p = 2 , q = 7 then 2 * 2 + 7 = 11 so 2 p + q = 11 . answer : a"
a ) 11 , b ) 14 , c ) 13 , d ) 15 , e ) 16
a
add(multiply(2, 2), 7)
multiply(n0,n2)|add(n1,#0)|
general
martin bought 10 concert tickets , some at the full price of $ 2.00 per ticket , and some at a discounted price of $ 1.60 per ticket . if he spent a total of $ 17.60 , how many discounted tickets did he buy ?
"let x be the number of tickets he bought at $ 2 per ticket . then 2 x + ( 10 - x ) 1.6 = 17.6 0.4 x = 1.6 = > x = 4 discounted tickets = 10 - x = 6 ans : d"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
d
subtract(multiply(10, 2.00), multiply(1.60, 10))
multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|
gain
the sum of the non - prime numbers between 50 and 60 , non - inclusive , is
sum of consecutive integers from 51 to 59 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 51 + 59 ) / 2 * 9 = 55 * 9 = 495 sum of non - prime numbers b / w 50 and 60 , non inclusive = = = > 495 - 112 ( i . e . , 53 + 59 , being the prime # s in the range ) = 383 answer : e
a ) 263 , b ) 293 , c ) 323 , d ) 353 , e ) 383
e
add(add(add(add(add(add(add(50, const_1), add(add(50, const_1), const_1)), add(add(add(50, const_1), const_1), const_2)), add(add(add(add(50, const_1), const_1), const_2), const_1)), add(add(add(add(add(50, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(50, const_1), const_1), const_2), cons...
add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)
general
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 500 square feet . how many feet offence does frank use ?
"area = length x breadth 500 = 40 x breadth so , breadth = 15 units fencing required is - breadth + breadth + length 15 + 15 + 40 = > 70 feet answer must be ( c ) 70"
a ) 14 , b ) 47 , c ) 70 , d ) 180 , e ) 240
c
add(add(divide(500, 40), divide(500, 40)), 40)
divide(n1,n0)|add(#0,#0)|add(n0,#1)|
geometry
the class mean score on a test was 80 , and the standard deviation was 15 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ?
"1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 95 to 65 , where 95 is within sd above the mean and 65 within 1 sd below the mean 2 sd = 15 twice = 30 from the the mean , which is 9...
a ) 30 , b ) 31 , c ) 50 , d ) 90 , e ) 89
c
multiply(2, 15)
multiply(n1,n2)|
general
there are 3 red chips and 3 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ?
"using anagram method : @nl 6 _ 5 _ 4 _ 3 _ 2 _ 1 r _ r _ r _ b _ b _ b so . . 6 ! / number of repeated letters ( 3 ! ) ( 3 ! ) = 20 ans : e"
a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 60 , e ) e ) 200
e
multiply(factorial(const_4.0), factorial(3))
factorial(const_4.0)|factorial(n1)|multiply(#0,#1)|
general
the average age of applicants for a new job is 32 , with a standard deviation of 8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the max...
"minimum age = average - 1 standard deviation = 32 - 8 = 24 maximum age = average + 1 standard deviation = 32 + 8 = 40 maximum number of different ages of the applicants = 30 - 24 + 1 = 7 answer c"
a ) 8 , b ) 77 , c ) 7 , d ) 18 , e ) 34
c
add(subtract(add(32, 8), subtract(32, 8)), const_1)
add(n0,n1)|subtract(n0,n1)|subtract(#0,#1)|add(#2,const_1)|
general
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number is :
"( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a"
a ) 36 , b ) 77 , c ) 88 , d ) 55 , e ) 221
a
add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5))))
multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
other
in a kilometer race , a beats b by 48 meters or 6 seconds . what time does a take to complete the race ?
"time taken by b run 1000 meters = ( 1000 * 6 ) / 48 = 125 sec . time taken by a = 125 - 6 = 119 sec . answer : b"
a ) 22 , b ) 119 , c ) 110 , d ) 109 , e ) 12
b
subtract(divide(multiply(const_1, const_1000), divide(48, 6)), 6)
divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|
physics
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 200 . find the total worth of the stock ?
"let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 200 = > x = 10000 answer : c"
a ) 20029 , b ) 20000 , c ) 10000 , d ) 20027 , e ) 20026
c
divide(200, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100))))
divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)|
gain
the average weight of 10 person ' s increases by 5 kg when a new person comes in place of one of them weighing 60 kg . what is the weight of the new person ?
"total increase in weight = 10 × 5 = 50 if x is the weight of the new person , total increase in weight = x − 60 = > 50 = x - 60 = > x = 50 + 60 = 110 answer : d"
a ) 85.5 , b ) 86.5 , c ) 87.5 , d ) 110 , e ) 89.5
d
add(multiply(10, 5), 60)
multiply(n0,n1)|add(n2,#0)|
general
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ?
"speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 60 / 5 = 12 seconds . answer : d"
a ) 22 seconds , b ) 65 seconds , c ) 78 seconds , d ) 12 seconds , e ) 21 seconds
d
divide(60, multiply(add(15, 3), const_0_2778))
add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
solution a is 10 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ?
"forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 -...
a ) 6 : 4 , b ) 6 : 14 , c ) 3 : 4 , d ) 4 : 6 , e ) 3 : 7
c
divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100)))))
divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)|
other
at what price must an article costing rs . 47.50 be marked in order that after deducting 12 % from the list price . it may be sold at a profit of 25 % on the cost price ?
"cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 88 / 100 ) = 59.375 mp = 67.5 answer : b"
a ) 62.5 , b ) 67.5 , c ) 62.7 , d ) 62.2 , e ) 62.9
b
divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 12))
multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)|
gain
if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a â ˆ ª b ) = 6 / 15 find p ( b | a )
p ( b | a ) = p ( a â ˆ ª b ) / p ( a ) p ( b | a ) = ( 6 / 15 ) / ( 2 / 15 ) = 3 . c
a ) 1 / 2 , b ) 2 / 3 , c ) 3 , d ) 4 / 5 , e ) 4 / 7
c
divide(divide(6, 15), divide(2, 15))
divide(n4,n1)|divide(n0,n1)|divide(#0,#1)
general
if x + y = 9 and x – y = 3 , then x ^ 2 - y ^ 2 =
the fastest approach has already been shown . here ' s one more option . given : x + y = 9 x – y = 3 add the two equations to get : 2 x = 12 , which means x = 6 if x = 6 , we can plug that value into either equation , to conclude that y = 3 if x = 6 and y = 3 , then x ² - y ² = 6 ² - 3 ² = 27 answer : d
a ) - 4 , b ) 4 , c ) 10 , d ) 27 , e ) 40
d
subtract(power(subtract(9, 3), 2), power(const_3, const_2))
power(const_3,const_2)|subtract(n0,n1)|power(#1,n2)|subtract(#2,#0)
general
in a primary school , the ratio of the number of girls to boys is 5 : 8 . if there are 160 girls , the total number of students in the college is :
let the number of girls and boys be 5 x and 8 x . total number of students = 13 x = 13 x 32 = 416 . answer : d
a ) 100 , b ) 250 , c ) 260 , d ) 300 , e ) 350
d
add(multiply(5, divide(160, 8)), add(160, multiply(5, 8)))
divide(n2,n1)|multiply(n0,n1)|add(n2,#1)|multiply(n0,#0)|add(#2,#3)
other
a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 35 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ?
"pastry is divided in 35 customers equally . so , total number of pastry must be a multiple of 35 only option a satisfies the condition , and hence is the answer"
a ) 525 . , b ) 412 . , c ) 432 . , d ) 502 . , e ) 522 .
a
multiply(divide(multiply(49, 6), subtract(49, 35)), 35)
multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)|
general
( 74 + 75 + 76 + 77 + . . . . . + 149 + 150 ) = ?
"explanation : sum of first n natural numbers = ( 1 + 2 + 3 + . . . . . + n ) = n ( n + 1 ) / 2 ( 74 + 75 + 76 + . . . . . + 150 ) = ( 1 + 2 + . . . . + 73 + 74 + 75 + . . . . . + 150 ) - ( 1 + 2 + . . . . + 73 ) = ( 1 / 2 x 150 x 151 ) - ( 1 / 2 x 73 x 74 ) = 11325 - 2701 = 8624 answer : option a"
a ) 8624 , b ) 8634 , c ) 8654 , d ) 8644 , e ) 8677
a
subtract(divide(multiply(77, add(77, const_1)), const_2), divide(multiply(subtract(74, const_1), 74), 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
an item is being sold for $ 10 each . however , if a customer will “ buy at least 3 ” they have a promo discount of 17 % . also , if a customer will “ buy at least 10 ” items they will deduct an additional 8 % to their “ buy at least 3 ” promo price . if sam buys 10 pcs of that item how much should he pay ?
"without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 27 % , thus he must pay slightly more than $ 73 . only answer choice e fits . answer : e"
a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 75.00
e
multiply(subtract(10, divide(multiply(17, 8), const_100)), 10)
multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,n0)|
gain
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 3 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ?
"1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 3 = 4320 . answer : e"
a ) 5729 , b ) 5760 , c ) 2889 , d ) 2870 , e ) 4320
e
divide(multiply(3, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1))
multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)|
physics
a feed store sells two varieties of birdseed : brand a , which is 65 % millet and 35 % sunflower , and brand b , which is 40 % millet and 60 % safflower . if a customer purchases a mix of the two types of birdseed that is 50 % millet , what percent of the mix is brand a ?
"yes there is a simple method : consider the following method brand a : 65 % millet and 35 % sunflower brand b : 40 % millet and 60 % safflower mix : 50 % millet here the weighted average is 50 % , now brand a has 65 % millet , which is 15 % more than the weighted average of mix = + 0.15 a - - - - - - - - - - - - - - -...
a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 55 %
a
multiply(divide(const_3, add(const_3, const_2)), const_100)
add(const_2,const_3)|divide(const_3,#0)|multiply(#1,const_100)|
gain
of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 16 boys and 44 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 w...
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
b
subtract(subtract(16, subtract(10, 4)), 6)
subtract(n2,n7)|subtract(n4,#0)|subtract(#1,n6)
other
pipe p can drain the liquid from a tank in 1 / 4 the time that it takes pipe q to drain it and in 2 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ?
"suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 1 / 4 ) rq ] = 4 also , rp = rr / ( 2 / 3 ) = > 4 = rr / ( 2 / 3 ) = > rr = 8 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 4 + 8 / 3 = 23 / 3 = 1 / ( 3 / 23 ) thus running simultaneo...
a ) 9 / 29 , b ) 3 / 23 , c ) 3 / 8 , d ) 17 / 29 , e ) 3 / 4
b
divide(multiply(1, 1), add(multiply(multiply(4, 1), const_2), multiply(1, 1)))
multiply(n0,n0)|multiply(n0,n1)|multiply(#1,const_2)|add(#2,#0)|divide(#0,#3)|
physics
a box contains 10 apples , 9 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ?
"when you choose ( and then eat ) the first apple , the probability of that apple being red is 9 / 10 . so if we do the activity 1 times , the probability of it being red is 9 / 10 . for 2 times , it is ( 9 / 10 ) * ( 8 / 9 ) for 3 times , it is ( 9 / 10 ) * ( 8 / 9 ) * ( 7 / 8 ) you can notice that the numerator of th...
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
d
add(multiply(0.5, const_10), const_1)
multiply(n2,const_10)|add(#0,const_1)|
general
find the circumference and area of radius 8 cm .
"area of circle = π r ² = 22 / 7 × 8 × 8 cm ² = 201 cm ² answer : b"
a ) 124 cm ² , b ) 201 cm ² , c ) 210 cm ² , d ) 184 cm ² , e ) 194 cm ²
b
circle_area(8)
circle_area(n0)|
geometry
if a - b = 10 and a 2 + b 2 = 210 , find the value of ab .
"2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 210 - 100 = 110 ab = 55 . answer : c"
a ) 68 , b ) 42 , c ) 55 , d ) 18 , e ) 44
c
divide(subtract(210, power(10, 2)), 2)
power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|
general
in a 150 member association consisting of men and women , exactly 10 % of men and exactly 20 % women are homeowners . what is the least number of members who are homeowners ?
solution simple out of 150 10 % are male i . e 15 and 20 % are female i . e 30 , so total homeowner is 45 . now min number homeowner is 15 and max is 30 so question ask us to find least and 18 has least value among all option . so ans is 18 . ans : e
a ) 22 , b ) 21 , c ) 20 , d ) 19 , e ) 18
e
add(multiply(multiply(divide(20, const_100), 10), multiply(divide(20, const_100), 10)), divide(subtract(150, 10), 10))
divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)
general
the length of the longest rod that can fit in a cubical room of 4 m side is ?
the length of the longest rod = 4 sq . root of 3 = 6.928 m answer : b
['a ) 6.978 m', 'b ) 6.928 m', 'c ) 6.829 m', 'd ) 6.248 m', 'e ) 6.982 m']
b
sqrt(add(power(multiply(sqrt(const_2), 4), const_2), power(4, const_2)))
power(n0,const_2)|sqrt(const_2)|multiply(n0,#1)|power(#2,const_2)|add(#3,#0)|sqrt(#4)
physics
if $ 5,000 is invested in an account that earns 16 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ?
"solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.16 / 2 ) ^ 2 = 5416 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 16 / 100 ) = 5400 difference = 5416 - 5400 = 16 $ d"
a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432
d
multiply(16, const_1)
multiply(n1,const_1)|
gain
a rectangular paper , when folded into two congruent parts had a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides . what is the area of the paper ?
solution when folded along breadth , we have : 2 ( l / 2 + b ) = 34 or l + 2 b = 34 when folded along length , we have : 2 ( l / 2 + b ) = 38 or 2 l + b = 38 solving ( i ) and ( ii ) , we get : l = 14 and b = 10 . ∴ area of the paper = ( 1410 ) cm 2 = 140 cm 2 answer a
['a ) 140 cm', 'b ) 240 cm 2', 'c ) 560 cm 2', 'd ) none of these', 'e ) can not be determined']
a
multiply(divide(subtract(multiply(38, const_2), 34), const_3), subtract(38, multiply(divide(subtract(multiply(38, const_2), 34), const_3), const_2)))
multiply(n1,const_2)|subtract(#0,n0)|divide(#1,const_3)|multiply(#2,const_2)|subtract(n1,#3)|multiply(#2,#4)
geometry
after giving a discount of rs . 50 the shopkeeper still gets a profit of 20 % , if the cost price is rs . 180 . find the markup % ?
"cost price = 180 s . p = 180 * 120 / 100 = 216 disc = 50 so . . . mark price = 216 + 50 = 266 . . . . . . mark up % = 266 - 180 / 180 = 86 / 180 = . 48 or 48 % answer : c"
a ) 0.5 , b ) 0.6 , c ) 0.48 , d ) 0.65 , e ) 0.45
c
divide(subtract(add(add(180, divide(180, divide(const_100, 20))), 50), 180), 180)
divide(const_100,n1)|divide(n2,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,n2)|divide(#4,n2)|
gain
the water level in a rectangular swimming pool measuring 60 feet by 10 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons )
"6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 60 * 10 * 1 / 2 = 300 feet ^ 3 of water must be removed , which equals to 300 * 7.5 = 2250 gallons . answer : b"
a ) 100 , b ) 2250 , c ) 750 , d ) 1200 , e ) 5625
b
multiply(volume_rectangular_prism(60, 10, divide(6, add(const_10, const_2))), 7.5)
add(const_10,const_2)|divide(n2,#0)|volume_rectangular_prism(n0,n1,#1)|multiply(n4,#2)|
geometry
a leak in the bottom of a tank can empty the tank in 6 hrs . an pipe fills water atthe rate of 4 ltrs / min . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hrs . the capacity of the tank is ?
1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 e
a ) 4670 , b ) 4546 , c ) 4556 , d ) 4675 , e ) 5760
e
multiply(multiply(multiply(6, 4), const_60), 4)
multiply(n0,n1)|multiply(#0,const_60)|multiply(n1,#1)
gain
what is the least number of squares tiles required to pave the floor of a room 2 m 25 cm long and 1 m 35 cm broad ?
"length of largest tile = h . c . f . of 225 cm and 135 cm = 45 cm . area of each tile = ( 45 x 45 ) cm 2 . required number of tiles = 225 x 135 / ( 45 ^ 2 ) = 15 . answer : a"
a ) 15 , b ) 20 , c ) 40 , d ) 44 , e ) 48
a
divide(rectangle_area(multiply(2, const_100), multiply(25, const_100)), square_area(add(multiply(const_4, const_10), const_1)))
multiply(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_1)|rectangle_area(#0,#1)|square_area(#3)|divide(#4,#5)|
physics
7 people average age is 30 . youngest person age is 8 . find average of the people when youngest was born .
"average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 56 = 154 so average age would be 154 / 7 = 22 answer : d"
a ) 23 , b ) 24 , c ) 25 , d ) 22 , e ) 27
d
divide(subtract(multiply(30, 7), multiply(8, 7)), 7)
multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n0)|
general
24 people can write 48 book in 6 days working 8 hour a day . then in how many day 192 can be written by 50 people ?
work per day epr hour per person = 48 / ( 6 * 8 * 24 ) / / eq - 1 people = 50 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 192 / ( p * 8 * 50 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 288 / 25 answer : e
a ) 223 / 26 , b ) 288 / 27 , c ) 309 / 22 , d ) 288 / 26 , e ) 288 / 25
e
divide(multiply(48, 6), divide(50, const_2))
divide(n5,const_2)|multiply(n1,n2)|divide(#1,#0)
physics
john makes $ 60 a week from his job . he earns a raise andnow makes $ 75 a week . what is the % increase ?
"increase = ( 15 / 60 ) * 100 = ( 1 / 4 ) * 100 = 25 % . b"
a ) 16.12 % , b ) 25 % , c ) 16.56 % , d ) 17.66 % , e ) 18.1 %
b
multiply(divide(subtract(75, 60), 60), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 5 on candy , how much did john spend at the supermarket ?
"let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t ....
a ) $ 75 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 180
a
divide(5, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2))))
divide(n0,n5)|divide(n0,n3)|divide(n0,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)|
general
a bus started its journey from chennai and reached mumbai in 50 min with its average speed of 60 km / hr . if the average speed of the bus is increased by 6 km / hr , how much time will it take to cover the same distance ?
sol . distance between mumbai and pune = ( 60 * 50 ) / 60 = 50 average speed of the bus is increased by 6 km / hr then the speed of the bus = 66 km / hr required time = 50 * 66 / 60 = 55 min . b
a ) 45 min . , b ) 55 min . , c ) 65 min . , d ) 60 min . , e ) 50 min .
b
multiply(divide(multiply(divide(50, 60), 60), subtract(60, 6)), 60)
divide(n0,n1)|subtract(n1,n2)|multiply(n1,#0)|divide(#2,#1)|multiply(n1,#3)
general
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ?
"relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 6 / 125 = 48 sec . answer : c"
a ) 22 sec , b ) 88 sec , c ) 48 sec , d ) 18 sec , e ) 9 sec
c
add(45, 30)
add(n1,n2)|
physics
the ratio between the present ages of a and b is 4.5 : 3 respectively . the ratio between a ' s age 4 years ago and b ' s age 4 years hence is 1 : 1 . what is the ratio between a ' s age 4 years hence and b ' s age 4 years ago ?
"let the present ages of a and b be 4.5 x and 3 x years respectively . then , ( 4.5 x - 4 ) / ( 3 x + 4 ) = 1 / 1 1.5 x = 8 = > x = 5.33 required ratio = ( 5 x + 4 ) : ( 3 x - 4 ) = 28 : 12 = 7 : 3 . answer : d"
a ) 3 : 4 , b ) 3 : 0 , c ) 3 : 1 , d ) 7 : 3 , e ) 3 : 7
d
divide(add(multiply(4.5, divide(add(4.5, 4.5), subtract(4.5, 3))), 4.5), subtract(multiply(3, divide(add(4.5, 4.5), subtract(4.5, 3))), 4.5))
add(n0,n0)|subtract(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n0,#3)|subtract(#4,n0)|divide(#5,#6)|
other
the average weight of a , b and c is 46 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is :
"let d sum of a , b , c is 3 * 46 = 144 and sum of a and b s 2 * 40 = 80 sum of b and c is 3 * 43 = 86 hence 80 + 86 - 144 = 22 ans = 22 answer : e"
a ) 17 kg , b ) 20 kg , c ) 26 kg , d ) 31 kg , e ) 22 kg
e
subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(46, const_3))
multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|
general
the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 6400 meters per second , what was the speed , in meters per second , when the stopwatch was started ?
"let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 6400 x = ( 2 ^ 8 * 25 ) / 2 ^ 9 = 25 / 2 the answer is a ."
a ) 25 / 2 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32
a
divide(divide(6400, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2)
divide(const_60,const_2)|add(#0,const_60)|divide(#1,n0)|subtract(#2,const_1)|power(const_2,#3)|divide(n1,#4)|divide(#5,const_2)|
physics
if two integers x , y ( x > y ) are selected from - 5 to 6 ( inclusive ) , how many cases are there ?
"there are 12 integers from - 5 to 6 inclusive . 12 c 2 = 66 . the answer is b ."
a ) 62 , b ) 66 , c ) 70 , d ) 74 , e ) 78
b
add(add(add(add(add(add(add(6, 5), add(6, const_2)), add(6, const_1)), 6), 5), const_2), const_1)
add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)|
general
what is the remainder when 2 ^ 20 is divided by 10 ?
"2 ^ 20 = 2 ^ 10 x 2 ^ 10 and 2 ^ 10 = 1024 4 * 4 = 16 , so the unit digit of this multiplication is 6 the remainder is 6 answer : d"
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
d
subtract(divide(10, const_2), multiply(2, 2))
divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)|
general
what will be the cost of building a fence around a square plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 56 ?
"let the side of the square plot be a ft . a 2 = 289 = > a = 17 length of the fence = perimeter of the plot = 4 a = 68 ft . cost of building the fence = 68 * 56 = rs . 3808 . answer : d"
a ) s . 3944 , b ) s . 3948 , c ) s . 3942 , d ) s . 3808 , e ) s . 3929
d
multiply(square_perimeter(sqrt(289)), 56)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
the average weight of a group of boys is 20 kg . after a boy of weight 34 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 34 kg joins the group , total weight of boys = 20 x + 34 so 20 x + 34 = 21 ( x + 1 ) = > x = 13 . answer : d"
a ) 12 , b ) 14 , c ) 18 , d ) 13 , e ) 10
d
add(subtract(34, add(20, 1)), 1)
add(n0,n2)|subtract(n1,#0)|add(#1,n2)|
general
in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 12 . what is the total number of students in the school ?
"explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 12 + 2 / 3 of 12 80 / 100 x = 20 x = 25 . answer : option c"
a ) 72 , b ) 80 , c ) 25 , d ) 150 , e ) 100
c
divide(add(12, multiply(12, divide(2, 3))), subtract(const_1, divide(20, const_100)))
divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3)|
general
if w is a positive integer and 10 ^ w – 74 in decimal notation has digits whose sum is 440 , what is the value of w ?
answer e = 50 we have ( 10 ^ w ) - 74 , where the sum of the digits is 440 . 100 - 74 = 26 , so we already have 2 + 6 = 8 from the 440 . because we have a power of 10 , we will have numbers like 100 , 1000 , 10000 and so on . this minus 74 rests 26 and a lot of 9 s . e . g : 1000 - 74 = 9926 . so dividing the 432 / 9 =...
a ) 40 , b ) 44 , c ) 45 , d ) 46 , e ) 50
e
add(subtract(74, divide(440, 10)), multiply(10, const_2))
divide(n2,n0)|multiply(n0,const_2)|subtract(n1,#0)|add(#1,#2)
general
the average age of 30 students of a class is 20 years . out of these , the average age of 10 students is 16 years and that of the other 19 students is 22 years . the age of the 30 th student is :
"explanation : age of the 30 th student = [ 30 x 20 - ( 10 x 16 + 19 x 22 ) ] = ( 600 - 278 ) = 22 years . answer : a"
a ) 22 years , b ) 24 years , c ) 15 years , d ) 16 years , e ) none of these
a
subtract(multiply(30, 30), add(multiply(10, 16), multiply(19, 22)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 4 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ?
"distance = relative speed * time = ( 4 + 3 ) * 4 = 28 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : a"
a ) 28 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 16 km
a
multiply(add(4, 3), 4)
add(n0,n1)|multiply(n2,#0)|
physics
in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 176 votes . find the total number of valid votes ?
"let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 176 = > 4 x / 10 = 176 = > 4 x = 1760 = > x = 440 . answer : a"
a ) 440 , b ) 570 , c ) 480 , d ) 520 , e ) 550
a
divide(176, divide(subtract(70, subtract(const_100, 70)), const_100))
subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|
gain
( 17 ) 5.25 x ( 17 ) ? = 178
"solution let ( 17 ) 5.25 * ( 17 ) x = 178 . then , ( 17 ) 5.25 + x = ( 17 ) 8 . ∴ 5.25 + x = 8 ⇔ x = ( 8 - 5.25 ) ⇔ x = 2.75 answer b"
a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these
b
multiply(17, 17)
multiply(n0,n2)|
general
in measuring the sides of a rectangle , one side is taken 20 % in excess and the other 10 % in deficit . find the error per cent in area calculated from the measurement ?
since side 1 x side 2 = area therefore , error % in area = ( x + y + xy / 100 ) % = [ 20 - 10 + ( - 10 x 20 ) / 100 ] % or 8 % i . e . , 8 % excess answer : c
['a ) 12 % deficit', 'b ) 10 % excess', 'c ) 8 % excess', 'd ) 9 % excess', 'e ) none of these']
c
subtract(subtract(20, 10), divide(multiply(20, 10), const_100))
multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)
geometry
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 54 kmph , find the speed of the stream ?
"the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 54 / 3 = 18 kmph . answer : b"
a ) 77 , b ) 18 , c ) 14 , d ) 12 , e ) 17
b
subtract(54, divide(multiply(54, const_2), const_3))
multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)|
physics
some ladies can do a piece of work in 12 days . two times the number of such ladies will do half of that work in :
expl : let x ladies can do the work in 12 days . more ladies , less days ( indirect ) less work , less days ( direct ) ladies 2 x : x work 1 : ½ 2 x : x , 1 : ½ : : 12 : y : . 2 x * 1 * y = x * ½ * 12 or y = 3 hence the required number of days = 3 answer : d
a ) 6 days , b ) 4 days , c ) 12 days , d ) 3 days , e ) 8 days
d
divide(12, const_4)
divide(n0,const_4)
physics
think of a number , divide it by 5 and add 6 to it . the result is 65 . what is the number thought of ?
"explanation : 65 - 6 = 59 59 x 5 = 295 answer : c"
a ) 24 , b ) 77 , c ) 295 , d ) 267 , e ) 29
c
multiply(subtract(65, 6), 5)
subtract(n2,n1)|multiply(n0,#0)|
general
sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 200 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every d...
"6 regular days - - > sales = 6 * 32 * x = 192 x ; 4 hot days - - > sales = 4 * 32 * ( 1.25 x ) = 160 x ; total sales = 192 x + 160 x = 352 x . total cost = 10 * 32 * 0.75 = 240 . profit = 352 x - 240 = 200 - - > x = 1.25 . 1.25 x = ~ 1.56 . answer : a ."
a ) $ 1.56 , b ) $ 1.88 , c ) $ 2.25 , d ) $ 2.50 , e ) $ 3.25
a
multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, 1), const_2)), 200), add(multiply(subtract(multiply(add(const_4, 1), const_2), 4), 32), multiply(multiply(divide(add(const_100, 25), const_100), 4), 32))), divide(add(const_100, 25), const_100))
add(n5,const_4)|add(n1,const_100)|divide(n2,const_100)|divide(#1,const_100)|multiply(n3,#2)|multiply(#0,const_2)|multiply(#4,#5)|multiply(n4,#3)|subtract(#5,n4)|add(n0,#6)|multiply(n3,#8)|multiply(n3,#7)|add(#10,#11)|divide(#9,#12)|multiply(#13,#3)|
gain
a candidate got 32 % of the votes polled and he lost to his rival by 1908 votes . how many votes were cast ?
"let x be the total number of votes . 0.32 x + 1908 = 0.68 x 0.36 x = 1908 x = 1908 / 0.36 = 5300 the answer is c ."
a ) 5100 , b ) 5200 , c ) 5300 , d ) 5400 , e ) 5500
c
divide(1908, subtract(subtract(const_1, divide(32, const_100)), divide(32, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|
gain
the sub - duplicate ratio of 1 : 9 is
"root ( 1 ) : root ( 9 ) = 1 : 3 answer : c"
a ) 4 : 3 , b ) 1 : 2 , c ) 1 : 3 , d ) 1 : 4 , e ) 2 : 3
c
divide(sqrt(1), sqrt(9))
sqrt(n0)|sqrt(n1)|divide(#0,#1)|
other
a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 482 more if the interest were payable half - yearly , then it were pay able yearly . find the sum .
"p ( 11 / 10 ) 4 - p ( 6 / 5 ) 2 = 482 p = 2000 answer : d"
a ) 3377 , b ) 6886 , c ) 2788 , d ) 2000 , e ) 28712
d
add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1))
add(n1,const_1)|multiply(#0,#0)|add(#0,#1)|
general
what is the value of 10 ^ 8 - 6 ^ 4 ?
"as 10 ^ n will always have last digit as 0 and 6 ^ n will always as last digit 6 . . hence difference of such sum should always be ending with 4 and there is only on option . . answer b"
a ) 9,99 , 98,623 , b ) 9,99 , 98,704 , c ) 9,99 , 98,800 , d ) 9,99 , 99,521 , e ) 9,99 , 99,932
b
divide(multiply(add(add(const_100, const_60), const_1), 10), const_100)
add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)|
general
what is the area of a square field whose diagonal of length 14 m ?
"d 2 / 2 = ( 14 * 14 ) / 2 = 140 answer : b"
a ) 160 sq m , b ) 140 sq m , c ) 200 sq m , d ) 400 sq m , e ) 180 sq m
b
divide(square_area(14), const_2)
square_area(n0)|divide(#0,const_2)|
geometry
harish can paint a wall in just 3 hours . however it takes 6 hours for ganpat to complete the same job of painting the wall . how long would it take for both of them to paint the wall , if both work together ?
d two hours . harish who could paint the wall in 3 hours could paint 2 / 3 of the wall in just two hours . ganpat who paint the wall in 6 hours could paint 1 / 3 of the wall in two hours . 2 / 3 + 1 / 3 = 1 .
a ) 3 hours , b ) 1 hours , c ) 3.20 hours , d ) 2 hours , e ) 1.35 hours
d
inverse(add(inverse(3), inverse(6)))
inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)
physics
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 38 years . what is deepak present age ?
"present age is 4 x and 3 x , = > 4 x + 6 = 38 = > x = 8 so deepak age is = 3 ( 8 ) = 24 answer : b"
a ) 22 , b ) 24 , c ) 77 , d ) 266 , e ) 182
b
divide(multiply(subtract(38, 6), 3), 4)
subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|
other
two pipes a and b can fill a tank in 40 and 120 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ?
"part filled by a in 1 min . = 1 / 40 part filled by b in 1 min . = 1 / 120 part filled by ( a + b ) in 1 min . = 1 / 40 + 1 / 120 = 1 / 30 . both the pipes can fill the tank in 30 minutes . answer : a"
a ) 30 , b ) 35 , c ) 25 , d ) 82 , e ) 22
a
add(multiply(40, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 40), divide(const_1, 120)), const_2)), 40), const_60))
divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)|
physics
rs 70000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 8000 find money given by 10 % ?
"let first parrt is x and second part is y then x + y = 70000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 8000 = ( x * 10 * 1 ) / 100 + ( y * 20 * 1 ) / 100 80000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 80000 = 70000 + y so y = 10000 then x = 70000 - 10...
a ) 30000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) 70000
d
divide(subtract(divide(multiply(70000, 20), const_100), 8000), divide(10, const_100))
divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(#2,n3)|divide(#3,#0)|
gain
in a two digit number , the digit in the units place is 6 times the digit in 10 ' s place and sum of the digits is equal to 7 . what is the number ?
solution let the ten digits be x . then , units digit = 6 x therefore x + 6 x = 7 ‹ = › 7 x = 7 ‹ = › x = 1 . so , ten ' s digit = 1 , unit ' s digit = 6 . hence , the required number is 16 . answer a
a ) 16 , b ) 28 , c ) 32 , d ) 36 , e ) 38
a
add(multiply(divide(7, add(6, const_1)), 10), multiply(6, divide(7, add(6, const_1))))
add(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|multiply(n0,#1)|add(#2,#3)
general
shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 40 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if...
"nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 38 hence answer is c"
a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46
c
divide(subtract(multiply(const_3, 40), 40), const_3)
multiply(n0,const_3)|subtract(#0,n0)|divide(#1,const_3)|
physics
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2340 , what will be b ' s share of the earnings ?
explanatory answer a , b and c will share the amount of $ 2340 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , th...
a ) $ 1100 , b ) $ 520 , c ) $ 780 , d ) $ 1170 , e ) $ 630
c
multiply(2340, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8)))))
inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)
physics
in an exam , amar scored 64 percent , bhavan scored 36 percent and chetan 44 percent . the maximum score awarded in the exam is 900 . find the average mark scored by all the 3 boys ?
average mark scored by all the three boys = [ 64 / 100 ( 900 ) + 36 / 100 ( 900 ) + 44 / 100 ( 900 ) ] / 3 = 432 answer : b
a ) 384 , b ) 432 , c ) 207 , d ) 269 , e ) 208
b
divide(add(multiply(divide(44, const_100), 900), add(multiply(divide(64, const_100), 900), multiply(divide(36, const_100), 900))), const_3)
divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n3,#0)|multiply(n3,#1)|multiply(n3,#2)|add(#3,#4)|add(#6,#5)|divide(#7,const_3)
general
a 280 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ?
"relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 280 ) / 9 = 500 / 9 x + 280 = 500 x = 220 . a"
a ) 220 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 330 m
a
subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 280)
add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)|
physics
find the average of first 5 multiples of 5 ?
"average = ( 5 + 10 + 15 + 20 + 25 ) / 5 = 15 answer is b"
a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 21
b
divide(add(add(add(5, const_1), add(add(5, const_1), const_2)), add(subtract(5, 5), subtract(5, const_2))), 5)
add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)|
general
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 108 km / hr . what is the length of the platform ?
"speed = 108 * 5 / 18 = 30 m / sec . length of the train = 30 * 20 = 600 m . let the length of the platform be x m . then , ( x + 600 ) / 36 = 30 = > x = 480 m . answer : d"
a ) 229 , b ) 240 , c ) 288 , d ) 480 , e ) 221
d
multiply(20, multiply(108, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics