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60 % of a number is added to 120 , the result is the same number . find the number ?
"( 60 / 100 ) * x + 120 = x 2 x = 600 x = 300 answer : a"
a ) 300 , b ) 288 , c ) 270 , d ) 129 , e ) 281
a
divide(120, divide(120, const_100))
divide(n1,const_100)|divide(n1,#0)|
gain
in a 60 member association consisting of men and women , exactly 20 % of men and exactly 25 % women are homeowners . what is the least number of members who are homeowners ?
"solution simple out of 60 20 % are male i . e 12 and 25 % are female i . e 15 , so total homeowner is 27 . now min number homeowner is 12 and max is 27 so question ask us to find least and 13 has least value among all option . so ans is 13 . answer : b"
a ) 29 , b ) 13 , c ) 25 , d ) 23 , e ) 21
b
add(multiply(multiply(divide(25, const_100), 20), multiply(divide(25, const_100), 20)), divide(subtract(60, 20), 20))
divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)|
general
if a card is drawn from a well shuffled pack of cards , the probability of drawing a spade or a king is -
explanation : p ( s Γ‘ Β΄ Ε“ k ) = p ( s ) + p ( k ) - p ( s Γ’ Λ† Β© k ) , where s denotes spade and k denotes king . p ( s Γ‘ Β΄ Ε“ k ) = 13 / 52 + 4 / 52 - 1 / 52 = 4 / 13 answer : d
a ) 8 / 13 , b ) 6 / 13 , c ) 3 / 13 , d ) 4 / 13 , e ) 1 / 13
d
add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52))
divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)
probability
a man swims downstream 30 km and upstream 18 km taking 6 hours each time , what is the speed of the man in still water ?
"30 - - - 6 ds = 5 ? - - - - 1 18 - - - - 6 us = 3 ? - - - - 1 m = ? m = ( 5 + 3 ) / 2 = 4 answer : e"
a ) 7 , b ) 8 , c ) 5 , d ) 2 , e ) 4
e
divide(add(divide(18, 6), divide(30, 6)), const_2)
divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|
physics
what is the least value of x . so that 23 x 59 is divisible by 3 .
"explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 3 + x + 5 + 9 = 19 + x least value of x may be 2 therefore 19 + 2 = 21 is divisible by 3 . answer : option a"
a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4
a
divide(divide(divide(lcm(23, 59), 59), const_4), const_4)
lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)|
general
if the sum of the 4 th term and the 12 th term of an arithmetic progression is 16 , what is the sum of the first 15 terms of the progression ?
"4 th term + 12 th term = 16 i . e . , ( a + 3 d ) + ( a + 11 d ) = 16 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 16 - - - - - - - - - - - - - - - from ( 1 ) = 120 answer : d"
a ) 80 , b ) 100 , c ) 105 , d ) 120 , e ) 110
d
multiply(divide(15, const_2), 16)
divide(n3,const_2)|multiply(n2,#0)|
general
the side of a square is increased by 25 % then how much % does its area increases ?
"a = 100 a 2 = 10000 a = 125 a 2 = 15625 - - - - - - - - - - - - - - - - 10000 - - - - - - - - - 5625 100 - - - - - - - ? = > 56.25 % answer : b"
a ) 56.28 , b ) 56.25 , c ) 56.2 , d ) 56.24 , e ) 56.21
b
divide(multiply(subtract(square_area(add(const_100, 25)), square_area(const_100)), const_100), square_area(const_100))
add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|
geometry
if x = 1 - 2 t and y = 2 t - 2 , then for what value of t does x = y ?
"we are given x = 1 – 2 t and y = 2 t – 2 , and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus , we can substitute 1 – 2 t for x and 2 t – 2 for y in the equation x = y . this gives us : 1 – 2 t = 2 t – 2 3 = 4 t 3 / 4 = t the answer is c ."
a ) 5 / 2 , b ) 3 / 2 , c ) 3 / 4 , d ) 2 / 5 , e ) 0
c
divide(add(1, 2), add(2, 2))
add(n0,n2)|add(n1,n2)|divide(#0,#1)|
general
pipes a and b can fill a tank in 6 and 4 hours . pipe c can empty it in 12 hours . if all pipes are opened together , then the tank will be filled in ?
net part filled in 1 hour = 1 / 6 + 1 / 4 - 1 / 12 = 1 / 3 the tank will be full in 3 hr answer is a
a ) 3 hr , b ) 2 hr , c ) 1 hr , d ) 9 / 7 hr , e ) 5 / 3 hr
a
inverse(subtract(add(divide(const_1, 6), divide(const_1, 4)), divide(const_1, 12)))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)
physics
hall is 15 m long and 12 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , the volume of the hall is :
2 ( 15 + 12 ) x h = 2 ( 15 x 12 ) h = 180 / 27 m = 20 / 3 m . volume = 15 x 12 x 20 / 3 m 3 answer : option c
['a ) 720', 'b ) 900', 'c ) 1200', 'd ) 1800', 'e ) 2100']
c
multiply(multiply(15, 12), divide(multiply(multiply(15, 12), const_2), multiply(add(15, 12), const_2)))
add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#0,const_2)|divide(#2,#3)|multiply(#4,#1)
geometry
a candidate got 35 % of the votes polled and he lost to his rival by 2370 votes . how many votes were cast ?
"35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2370 100 % - - - - - - - - - ? = > 7900 answer : b"
a ) 7500 , b ) 7900 , c ) 2665 , d ) 2888 , e ) 2661
b
divide(2370, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100)))
divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)|
gain
sonika bought a v . c . r . at the list price of 18,500 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r .
sol . list price of v . c . r . = 18,500 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,500 = 8 ⁄ 100 Γ— 18500 = 1480 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,500 + 1480 = 19,980 . answer a
a ) 19,980 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these
a
floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2)))
add(const_1,const_4)|add(n1,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|add(n1,#2)|multiply(#3,const_2)|multiply(#4,const_100)|multiply(#6,#2)|add(#7,#3)|multiply(#8,#1)|divide(#9,const_100)|divide(#10,#5)|floor(#11)
gain
a cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the west surface is :
"solution area of the west surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 Γ— 1.25 + 6 Γ— 1.25 ) + 6 Γ— 4 ] m 2 = 49 m 2 . answer a"
a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 55 m 2 , e ) none of these
a
add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 6))), multiply(4, 6))
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
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 220 ?
"i can buy 9 250 - pack for rs 22.95 * 9 = $ 206.55 now , i can buy 5 20 - pack for 3.05 * 5 = $ 15.25 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 5 = 2218 e"
a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,124 , e ) 2,218
e
multiply(divide(220, 22.95), 250)
divide(n6,n5)|multiply(n4,#0)|
general
if ( n + 2 ) ! / n ! = 90 , n = ?
"( n + 2 ) ! / n ! = 90 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 132 from here , we might just test the answer choices . since ( 10 ) ( 9 ) = 90 , we can see that n = 8 b"
a ) 2 / 131 , b ) 8 , c ) 10 , d ) 11 , e ) 12
b
subtract(add(const_4, const_4), const_1)
add(const_4,const_4)|subtract(#0,const_1)|
general
abcd is a square . f and e are the midpoints of sides ad and cd , respectively . the area of triangle fed is 2 square inches . what is the area of square abcd ( in square inches ) ?
area of fed = 2 sq inches = 1 / 2 * de * fd = 1 / 2 * de ^ 2 because the sides of a square are equal , hence half of the sides will also be equal . de ^ 2 = 4 de = fd = 2 hence the side of the square = 4 area if the square = 4 * 4 = 16 correct option : c
['a ) 4', 'b ) 8', 'c ) 16', 'd ) 32', 'e ) 64']
c
multiply(multiply(2, const_2), const_4)
multiply(n0,const_2)|multiply(#0,const_4)
geometry
the triplicate ratio of 1 : 7 is ?
"1 ^ 3 : 7 ^ 3 = 1 : 343 answer : c"
a ) 1 : 7 , b ) 1 : 8 , c ) 1 : 343 , d ) 1 : 1 , e ) 1 : 2
c
divide(power(1, 7), power(7, 7))
power(n0,n1)|power(n1,n1)|divide(#0,#1)|
other
what profit percent is made by selling an article at a certain price , if by selling at 3 / 4 rd of that price , there would be a loss of 20 % ?
"sp 2 = 3 / 4 sp 1 cp = 100 sp 2 = 80 3 / 4 sp 1 = 80 sp 1 = 107 100 - - - 107 = > 7 % answer : c"
a ) 20 % , b ) 29 % , c ) 7 % , d ) 27 % , e ) 28 %
c
subtract(divide(subtract(const_100, 20), divide(3, 4)), const_100)
divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|
gain
275.124 x 16.98 Γ£ Β· 5.001 + 22.22 = ?
"explanation : ? = 275.124 x 16.98 Γ£ Β· 5.001 + 22.22 = ? Γ’ ‰ Λ† ( 275.124 x 17 / 5 ) + 22.22 Γ’ ‰ Λ† 935.421 + 22.22 Γ’ ‰ Λ† 957.641 answer : option c"
a ) 983.578 , b ) 659.121 , c ) 957.641 , d ) 656.112 , e ) 456.512
c
multiply(275.124, power(16.98, 5.001))
power(n1,n2)|multiply(n0,#0)|
general
a clock store sold a certain clock to a collector for 30 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 60 percent of what the collector had paid . the shop then sold the clock again at a profit of 90 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 150 , for how much did the shop sell the clock the second time ?
"now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 30 % profit . this means the clocks ' selling price was c ( 1.3 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 60 % the price at which the collector bought it . thus , you get = 1.3 * 0.6 * c = 0.78 c furthermore , the store sells the clock for the second time for 90 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.9 = 1.9 * 0.78 c finally given that c - 0.78 c = 150 - - - - > c = 681.82 $ thus , the cost of the clock the second time around = 1.9 * 0.78 c = 1.9 * 0.78 * 681.82 = 1010.46 $ . hence a is the correct answer ."
a ) $ 1010.46 , b ) $ 505.23 , c ) $ 2020.92 , d ) $ 800.46 , e ) $ 1515.69
a
divide(multiply(150, divide(multiply(add(150, 90), divide(add(150, 30), const_2)), 150)), subtract(divide(add(150, 30), const_2), 30))
add(n2,n3)|add(n0,n3)|divide(#1,const_2)|multiply(#0,#2)|subtract(#2,n0)|divide(#3,n3)|multiply(n3,#5)|divide(#6,#4)|
general
a certain junior class has 1000 students and a certain senior class has 900 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ?
"let ' s see pick 60 / 1000 first then we can only pick 1 other pair from the 800 so total will be 60 / 900 * 1000 simplify and you get 2 / 30000 answer is b"
a ) 3 / 40000 , b ) 2 / 30000 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15
b
divide(1, const_3)
divide(n3,const_3)|
probability
a money lender lent rs . 1000 at 3 % per year and rs . 1400 at 3 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years .
( 1000 xtx 3 / 100 ) + ( 1400 xtx 3 / 100 ) = 350 Γ’ † ’ t = 4.86 answer e
a ) 3.5 , b ) 3.75 , c ) 4 , d ) 4.25 , e ) 4.86
e
divide(350, add(divide(multiply(3, 1000), const_100), divide(multiply(1400, 3), const_100)))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|divide(n4,#4)|
gain
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 400 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ?
"i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 400 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 400 ) / 100 + ( 1.8 * 700 ) / 100 = 15.8 gms % of liquid x in resultant mixture = ( 15.8 / 1000 ) * 100 = 1.58 % a"
a ) 1.58 % , b ) 1.98 % , c ) 10 % , d ) 15 % , e ) 19 %
a
divide(add(multiply(400, 0.8), multiply(700, 1.8)), const_1000)
multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)|
gain
if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 21 , find the value of ab .
"2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 21 - 9 = 12 ab = 6 . answer is d ."
a ) 5 , b ) 8 , c ) 4 , d ) 6 , e ) 3
d
divide(subtract(21, power(3, 2)), 2)
power(n0,n1)|subtract(n3,#0)|divide(#1,n1)|
general
out of 460 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ?
"n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 460 - 50 = 410 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 410 = 90 . answer is e"
a ) 120 , b ) 150 , c ) 100 , d ) 180 , e ) 90
e
subtract(add(175, 325), subtract(460, 50))
add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|
other
in a 8 x 8 chess board what is the total number of squares .
ans : the total number of squares in a n x n chess board is equal to ` ` the sum of first n natural number squares ' ' i . e . , n ( n + 1 ) ( 2 n + 1 ) 6 n ( n + 1 ) ( 2 n + 1 ) 6 so substituting 8 in the above formula we get 204 answer : b
a ) 238 , b ) 204 , c ) 678 , d ) 169 , e ) 161
b
divide(multiply(multiply(8, add(8, const_1)), add(multiply(8, const_2), const_1)), subtract(8, const_2))
add(n0,const_1)|multiply(n0,const_2)|subtract(n0,const_2)|add(#1,const_1)|multiply(n0,#0)|multiply(#3,#4)|divide(#5,#2)
general
in a division , divident is 686 , divisior is 36 and quotient is 19 . find the remainder .
"explanation : 686 = 36 x 19 + r 686 = 684 + r r = 686 - 684 = 2 answer : option c"
a ) a ) 4 , b ) b ) 3 , c ) c ) 2 , d ) d ) 5 , e ) e ) 6
c
reminder(686, 36)
reminder(n0,n1)|
general
a can do a work in 15 days and b in 20 days . if they work on it together for 4 days , then the fraction of the work that is left is :
"explanation : a ' s 1 day ' s work = 1 / 15 b ' s 1 day ' s work = 1 / 20 ( a + b ) ' s 1 day ' s work = ( 1 / 15 + 1 / 20 ) = 7 / 60 ( a + b ) ' s 4 day ' s work = 7 / 60 x 4 = 7 / 15 therefore , remaining work = 1 - 7 / 15 = 8 / 15 answer is d"
a ) 1 / 4 , b ) 1 / 10 , c ) 7 / 15 , d ) 8 / 15 , e ) 9 / 15
d
subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 4))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|
physics
rakesh credits 15 % of his salary in his fixed deposit account and spends 30 % of the remaining amount on groceries . if the cash in hand is rs . 2380 , what is his salary ?
explanation : let salary be rs . x . then , x - 15 % of x - 30 % of 85 % of x = 2380 or x - 15 x / 100 βˆ’ 30 Γ— 85 Γ— x / 100 Γ— 100 = 2380 or 200 x - 30 x - 51 x = 2380 Γ— 2002380 Γ— 200 or 119 x = 2380 Γ— 2002380 Γ— 200 or x 2380 Γ— 200 / 119 = 4000 correct option : b
a ) rs . 3500 , b ) rs . 4000 , c ) rs . 4500 , d ) rs . 5000 , e ) none
b
divide(2380, multiply(divide(subtract(const_100, 15), const_100), divide(subtract(const_100, 30), const_100)))
subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)
gain
a salesman Γ’ € β„’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth
"[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 12000 answer a"
a ) 12000 , b ) 14000 , c ) 15000 , d ) 20000 , e ) 60000
a
subtract(multiply(5, const_4), const_12)
multiply(n0,const_4)|subtract(#0,const_12)|
general
if a light flashes every 6 seconds , how many times will it flash in ΒΎ of an hour ?
there are 60 minutes in an hour . in ΒΎ of an hour there are ( 60 * ΒΎ ) minutes = 45 minutes . in ΒΎ of an hour there are ( 60 * 45 ) seconds = 2700 seconds . light flashed for every 6 seconds . in 2700 seconds 2700 / 6 = 450 times . the count start after the first flash , the light will flashes 451 times in ΒΎ of an hour . answer : a
a ) 451 times , b ) 638 times , c ) 838 times , d ) 436 times , e ) 435 times
a
divide(multiply(divide(const_3600, const_4), const_3), 6)
divide(const_3600,const_4)|multiply(#0,const_3)|divide(#1,n0)
physics
by selling an article at rs . 400 , a profit of 60 % is made . find its cost price ?
"sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 60 ) ] = 400 * [ 100 / 160 ] = rs . 250 answer : d"
a ) 228 , b ) 267 , c ) 287 , d ) 250 , e ) 811
d
divide(multiply(400, const_100), add(const_100, 60))
add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|
gain
while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 3 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ?
"both computers together process files at a rate of 240 / 12 + 240 / 3 = 20 + 80 = 100 files per hour . the time required to process 240 files is 240 / 100 = 2.4 hours the answer is e ."
a ) 1.6 , b ) 1.8 , c ) 2 , d ) 2.2 , e ) 2.4
e
divide(240, add(divide(240, 12), divide(240, 3)))
divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(n0,#2)|
physics
on dividing a number by 357 , we get 39 as remainder . on dividing the same number 17 , what will be the remainder ?
"explanation : let x be the number and y be the quotient . then , x = 357 x y + 39 = ( 17 x 21 x y ) + ( 17 x 2 ) + 5 = 17 x ( 21 y + 2 ) + 5 ) required remainder = 5 . c )"
a ) 10 , b ) 12 , c ) 5 , d ) 25 , e ) 85
c
multiply(subtract(divide(power(39, const_2), 357), floor(divide(power(39, const_2), 357))), 357)
power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)|
general
two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team r . if team a ' s average speed was 5 mph greater than team r ' s , what was team r ' s average mph ?
"this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team r ' s speed was s . then team r ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team r . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team r is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct ."
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25
d
divide(divide(300, 5), 3)
divide(n0,n2)|divide(#0,n1)|
physics
the average marks of a class of 24 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ?
"sum of the marks for the class of 24 students = 24 * 40 = 960 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 74 students = 960 + 3000 = 3960 average marks of all the students = 3960 / 74 = 53.5 answer : d"
a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 53.5 , e ) 52.1
d
divide(add(multiply(24, 40), multiply(50, 60)), add(24, 50))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|
general
each month a retailer sells 100 identical items . on each item he makes a profit of $ 30 that constitutes 10 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ?
"for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 10 % of the cost : sell price = cost + $ 30 sell price = $ 300 + $ 30 thus , the sell price is $ 330 for each item . selling all 100 items gives the retailer . . . 100 ( $ 30 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 330 ) = $ 16.5 discount $ 313.5 = $ 300 + $ 13.5 now , the retailer makes a profit of just $ 13.5 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 13.5 ( x ) = $ 2,000 x = 2,000 / 13.5 x = 222.222222 items you ' ll notice that this is not among the answer choices . . . . 221 and 223 are . selling 221 items would get us 9 ( 221 ) = $ 1989 which is not enough money . to get back to at least $ 2,000 , we need to sell 223 items . final answer : d"
a ) 191 , b ) 213 , c ) 221 , d ) 223 , e ) 226
d
divide(multiply(100, 30), subtract(30, divide(multiply(add(divide(multiply(100, 30), 10), 30), 5), 100)))
multiply(n0,n1)|divide(#0,n2)|add(n1,#1)|multiply(n3,#2)|divide(#3,n0)|subtract(n1,#4)|divide(#0,#5)|
gain
if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 555 will be :
"z share = z , y = 1.2 z x = 1.25 Γ£ β€” 1.2 z , x + y + z = 555 ( 1.25 Γ£ β€” 1.2 + 1.2 + 1 ) z = 55 3.7 z = 555 , z = 150 answer : . d"
a ) rs . 300 , b ) rs . 200 , c ) rs . 240 , d ) rs . 150 , e ) none of these
d
divide(555, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1))
divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,#2)|add(#2,#4)|add(#5,const_1)|divide(n2,#6)|
general
for integers x , y , and z , if ( 3 ^ x ) ( 4 ^ y ) ( 5 ^ z ) = 13107 , 200000 and x + y + z = 16 , what is the value of xy / z ?
the number 13,107 , 200,000 is not divisible by 3 . ( we can verify this by adding the digits of the number to see that the sum is not a multiple of 3 . ) thus x = 0 and xy / z = 0 . the answer is b .
a ) undefined , b ) 0 , c ) 3 , d ) 5 , e ) 7
b
subtract(4, 3)
subtract(n1,n0)
general
the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 972 sq m , then what is the breadth of the rectangular plot ?
"let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 972 3 b 2 = 972 b 2 = 324 b = 18 m . answer : option c"
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 14
c
sqrt(divide(972, const_3))
divide(n0,const_3)|sqrt(#0)|
geometry
the mean of 15 observations was 25 . it was found later that an observation 40 was wrongly taken as 25 . the corrected new mean is
"explanation : correct sum = ( 25 * 15 + 40 - 25 ) = 390 correct mean = = 390 / 15 = 26 answer : e"
a ) 20 , b ) 30 , c ) 15 , d ) 18 , e ) 26
e
divide(add(multiply(25, 15), subtract(subtract(15, const_2), 25)), 15)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
what is the area of a square field whose diagonal of length 24 m ?
"d 2 / 2 = ( 24 * 24 ) / 2 = 240 answer : a"
a ) 240 sq m , b ) 250 sq m , c ) 200 sq m , d ) 400 sq m , e ) 800 sq m
a
divide(square_area(24), const_2)
square_area(n0)|divide(#0,const_2)|
geometry
if 40 % of a number is equal to two - thirds of another number , what is the ratio of the first number to the second ?
"suppose the first number is x and the second numbery . therefore , 40 % of x = 2 ⁄ 3 of y ∴ x ⁄ y = 2 ⁄ 3 Γ— 100 ⁄ 40 = 5 ⁄ 3 answer d"
a ) 7 : 3 , b ) 3 : 7 , c ) 2 : 5 , d ) 5 : 3 , e ) none of these
d
divide(divide(const_1, const_4), divide(40, const_100))
divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)|
general
( x ) + 3612 + 16125 - 6149 = 90189 . calculate the value of x
"x + 3612 + 16125 - 6149 = 90189 = x + 3612 + 16125 = 90189 + 6149 = x + 19737 = 96338 = x = 96338 - 19737 = 76601 answer is a"
a ) 76601 , b ) 76600 , c ) 76655 , d ) 76313 , e ) 76723
a
multiply(subtract(subtract(add(90189, 6149), 16125), 3612), divide(const_60, const_2))
add(n2,n3)|divide(const_60,const_2)|subtract(#0,n1)|subtract(#2,n0)|multiply(#1,#3)|
general
if the radius of a circle is increased by 12 % , then the area of the circle
a 1 = pi ( 100 ) ^ 2 a 2 = pi ( 112 ) ^ 2 so ( a 2 - a 1 ) / a 1 * 100 = 25.44 answer : b
['a ) decreases by 25.44 %', 'b ) increases by 25.44 %', 'c ) no change in area', 'd ) decreases by 12 %', 'e ) none']
b
multiply(subtract(power(add(divide(12, const_100), const_1), const_2), const_1), const_100)
divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)
geometry
a grocer has a sale of rs . 2500 , rs . 6500 , rs . 9855 , rs . 7230 and rs . 7000 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 7500 ?
total sale for 5 months = rs . ( 2500 + 6500 + 9855 + 7230 + 7000 ) = rs . 33085 . required sale = rs . [ ( 7500 x 6 ) - 33085 ] = rs . ( 45000 - 33085 ) = rs . 11915 . c
a ) s . 49180 , b ) s . 49910 , c ) s . 11915 , d ) s 6997 , e ) s . 5000
c
subtract(multiply(add(5, const_1), 7500), add(add(add(add(2500, 6500), 9855), 7230), 7000))
add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)
general
a cyclist bikes x distance at 20 miles per hour and returns over the same path at 16 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ?
"distance = d 1 = x miles speed = s 1 = 20 miles per hour time = t 1 = distance / speed = x / 20 2 . going from b to a distance = d 2 = x miles speed = s 2 = 16 miles per hour time = t 2 = distance / speed = x / 16 3 . average speed = total distance / total time total distance = x + x = 2 x total time = x / 20 + x / 16 = 9 x / 80 speed = 2 x / ( 9 x / 80 ) = 160 / 9 = 17.6 answer : a"
a ) 17.6 , b ) 17.3 , c ) 8.6 , d ) 17.2 , e ) 9.0
a
divide(add(20, 16), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
if a is an integer greater than 6 but less than 17 and b is an integer greater than 3 but less than 29 , what is the range of a / b ?
"the way to approach this problem is 6 < a < 17 and 3 < b < 29 minimum possible value of a is 7 and maximum is 16 minimum possible value of b is 4 and maximum is 28 range = max a / min b - min a / max b ( highest - lowest ) 16 / 4 - 7 / 28 = 15 / 4 hence a"
a ) 15 / 4 , b ) 13 / 2 , c ) 9 / 7 , d ) 1 / 5 , e ) 7 / 6
a
subtract(divide(subtract(17, const_1), add(3, const_1)), divide(add(6, const_1), subtract(29, 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
increasing the original price of a certain item by 25 percent and then increasing the new price by 25 percent is equivalent to increasing the original price by what percent ?
"soln : - x * 1.25 * 1.25 = 1.5625 x so there is a net increase of 56.25 % . answer e ."
a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 56.25
e
multiply(subtract(multiply(add(divide(25, const_100), const_1), add(divide(25, const_100), const_1)), const_1), const_100)
divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)|
gain
ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 61900 . what is his profit percent ?
"total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 61900 profit ( % ) = ( 61900 - 55000 ) / 55000 * 100 = 12.5 % answer : e"
a ) 22 , b ) 77 , c ) 18 , d ) 99 , e ) 12.5
e
multiply(divide(subtract(61900, add(42000, 13000)), add(42000, 13000)), const_100)
add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|
gain
increasing the original price of an article by 5 percent and then increasing the new price by 5 percent is equivalent to increasing the original price by
"1.05 * 1.05 * x = 1.1025 * x the answer is a ."
a ) 10.25 % , b ) 12.5 % , c ) 14.75 % , d ) 15.15 % , e ) 16.25 %
a
multiply(const_100, subtract(multiply(add(const_1, divide(5, const_100)), add(const_1, divide(5, const_100))), const_1))
divide(n0,const_100)|add(#0,const_1)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)|
gain
there are 1200 jelly beans divided between two jars , jar x and jar y . if there are 400 fewer jelly beans in jar x than 3 times the number of beans in jar y , how many beans are in jar x ?
x + y = 1200 so y = 1200 - x x = 3 y - 400 x = 3 ( 1200 - x ) - 400 4 x = 3200 x = 800 the answer is d .
a ) 650 , b ) 700 , c ) 750 , d ) 800 , e ) 850
d
subtract(1200, divide(add(1200, 400), const_4))
add(n0,n1)|divide(#0,const_4)|subtract(n0,#1)
general
if x is 20 percent more than y and y is 70 percent less than z , then x is what percent of z ?
z = 100 ; y = 30 so x = 36 x as % of z = 36 / 100 * 100 = > 36 % answer will be ( d )
a ) 500 % , b ) 250 % , c ) 500 / 3 % , d ) 36 % , e ) 60 %
d
multiply(add(const_1, divide(20, const_100)), subtract(const_100, 70))
divide(n0,const_100)|subtract(const_100,n1)|add(#0,const_1)|multiply(#2,#1)
general
how many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language ?
1 million distinct 3 digit initials are needed . let the number of required alphabets in the language be β€˜ n ’ . therefore , using β€˜ n ’ alphabets we can form n * n * n = n 3 distinct 3 digit initials . note distinct initials is different from initials where the digits are different . for instance , aaa and bbb are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different . this n 3 different initials = 1 million i . e . n 3 = 106 ( 1 million = 106 ) = > n 3 = ( 102 ) 3 = > n = 102 = 100 hence , the language needs to have a minimum of 100 alphabets to achieve the objective . ans : d
a ) 321 , b ) 65 , c ) 120 , d ) 100 , e ) 80
d
divide(multiply(multiply(1, const_1000), const_1000), multiply(const_10, const_1000))
multiply(n0,const_1000)|multiply(const_10,const_1000)|multiply(#0,const_1000)|divide(#2,#1)
general
a train covers a distance of 12 km in 10 min . if it takes 8 sec to pass a telegraph post , then the length of the train is ?
"speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 8 = 160 m . answer : option c"
a ) 100 , b ) 110 , c ) 160 , d ) 130 , e ) 140
c
divide(12, subtract(divide(12, 10), 8))
divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)|
physics
at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 11 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 3 to 14 . how many cooks does the restaurant have ?
"originally there were 3 k cooks and 11 k waiters . 14 k = 11 k + 12 k = 4 there are 12 cooks . the answer is d ."
a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15
d
multiply(divide(12, 14), divide(12, 14))
divide(n2,n4)|multiply(#0,#0)|
other
bruce purchased 8 kg of grapes at the rate of 70 per kg and 11 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ?
"cost of 8 kg grapes = 70 Γ— 8 = 560 . cost of 11 kg of mangoes = 55 Γ— 11 = 605 . total cost he has to pay = 560 + 605 = 1165 b"
a ) a ) 1040 , b ) b ) 1165 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1075
b
add(multiply(8, 70), multiply(11, 55))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
gain
a 56 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ?
"amount of salt = 5.6 assume x gallons of water are added . 5.6 / 56 + x = 8 / 100 560 = 8 x + 448 8 x = 112 x = 14 correct option : d"
a ) 8 , b ) 12 , c ) 13 , d ) 14 , e ) 16
d
divide(multiply(56, subtract(10, 8)), 8)
subtract(n1,n2)|multiply(n0,#0)|divide(#1,n2)|
general
a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 120 more than a :
"explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 120 total earning = \ inline \ frac { 58 xy } { 100 } = 3480 answer : b ) rs . 3480"
a ) 2348 , b ) 3480 , c ) 2767 , d ) 1998 , e ) 2771
b
multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(120, subtract(multiply(4, 5), multiply(3, 6))))
multiply(n0,n3)|multiply(n1,n2)|add(#0,#1)|subtract(#1,#0)|add(#2,#1)|divide(n6,#3)|multiply(#4,#5)|
general
if log 2 = 0.3010 and log 3 = 0.4771 , the value of log 5 ( 512 )
"log 5 ( 512 ) = log ( 512 ) / log 5 = log 2 ^ 9 / log ( 10 / 2 ) = 9 log 2 / ( log 10 - log 2 ) = ( 9 x 0.3010 ) / ( 1 - 0.3010 ) = 2.709 / 0.699 = 2709 / 699 = 3.876 answer is a ."
a ) 3.876 , b ) 2.967 , c ) 2.87 , d ) 3.912 , e ) 1.9
a
multiply(0.3010, divide(0.3010, 0.4771))
divide(n1,n3)|multiply(n1,#0)|
other
what is the smallest positive integer x such that 507 + x is the cube of a positive integer ?
"given 507 + x is a perfect cube so we will take 512 = 8 * 8 * 8 507 + x = 512 x = 512 - 507 = 5 correct option is b"
a ) 15 , b ) 5 , c ) 50 , d ) 2 , e ) 4
b
add(const_3, const_4)
add(const_3,const_4)|
general
a train which has 440 m long , is running 45 kmph . in what time will it cross a person moving at 9 kmph in same direction ?
"time taken to cross a moving person = length of train / relative speed time taken = 440 / ( ( 45 - 9 ) ( 5 / 18 ) = 440 / 36 * ( 5 / 18 ) = 440 / 10 = 44 sec answer : b"
a ) 56 sec , b ) 44 sec , c ) 36 sec , d ) 29 sec . , e ) 19 sec .
b
divide(440, subtract(divide(45, const_3_6), divide(divide(9, const_2), const_3_6)))
divide(n1,const_3_6)|divide(n2,const_2)|divide(#1,const_3_6)|subtract(#0,#2)|divide(n0,#3)|
physics
two numbers have a h . c . f of 16 and a product of two numbers is 2560 . find the l . c . m of the two numbers ?
"l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 2560 / 16 = 160 . answer : c"
a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 180
c
divide(2560, 16)
divide(n1,n0)|
physics
what is the tens digit of 36 ^ 4 ?
"36 ^ 10 = 6 ^ 20 ( 6 ^ 2 ) = 6 * 6 = 36 ( 6 ^ 3 ) = 36 * 6 = . 16 ( 6 ^ 4 ) = . 16 * 6 = . . 96 ( 6 ^ 5 ) = . . 96 * 6 = . . 76 ( 6 ^ 6 ) = . . 76 * 6 = . . . 56 ( 6 ^ 7 ) = . . . . 56 * 6 = . . . . 36 if you see there is a pattern here in tens digits 3 , 1,9 , 7,5 , 3,1 and so on . . . continue the pattern up to 6 ^ 6 ( dont actually calculate full values ) and answer is c : 5"
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9
c
floor(divide(reminder(power(36, reminder(4, add(const_4, const_1))), const_100), const_10))
add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)|
general
in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 38 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ?
"girls = 900 , boys = 1100 passed = ( 38 % of 1100 ) + ( 32 % of 900 ) = 418 + 288 = 706 failed = 2000 - 706 = 1294 failed % = [ ( 1294 / 2000 ) x 100 ] % = 64.7 % . answer : b"
a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 69.57 % , e ) none of these
b
multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(38, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100)
divide(n2,const_100)|divide(n3,const_100)|multiply(const_10,const_100)|multiply(#2,const_2)|multiply(n1,#1)|subtract(#3,n1)|multiply(#0,#5)|subtract(#3,#6)|subtract(#7,#4)|divide(#8,#3)|multiply(#9,const_100)|
general
an article is bought for rs . 600 and sold for rs . 450 , find the loss percent ?
"600 - - - - 150 100 - - - - ? = > 16 2 / 3 % answer : e"
a ) 16 % , b ) 18 % , c ) 19 % , d ) 20 % , e ) 25 %
e
subtract(const_100, divide(multiply(450, const_100), 600))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|
gain
0.003 * 0.5 = ?
"explanation : 3 * 5 = 15 . sum of decimal places = 4 . 0.003 * 0.5 = 0.0015 = 0.0015 answer b"
a ) 0.00015 , b ) 0.0015 , c ) 0.015 , d ) 0.15 , e ) none of these
b
multiply(divide(0.003, 0.5), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
a man rows his boat 60 km downstream and 30 km upstream taking 3 hrs each time . find the speed of the stream ?
explanation : speed of the boat downstream = speed of the boat upstream \ small \ therefore the speed of the stream = answer : a
a ) 5 kmph , b ) 6 kmph , c ) 8 kmph , d ) 1 kmph , e ) 2 kmph
a
divide(subtract(divide(60, 3), divide(30, 3)), const_2)
divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)
physics
the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 3 : 2 . if the oil bill for february had been $ 10 more , the corresponding ratio would have been 5 : 3 . how much was the oil bill for january ?
"3 : 2 = 9 : 6 and 5 : 3 = 10 : 6 . an increase in $ 10 increases the ratio by 1 : 6 . therefore , january ' s bill was 6 ( $ 10 ) = $ 60 . the answer is a ."
a ) $ 60 , b ) $ 80 , c ) $ 100 , d ) $ 120 , e ) $ 140
a
divide(10, subtract(divide(5, 3), divide(3, 2)))
divide(n3,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2)|
general
what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 600 inclusive ?
"this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 600 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 60 ( 10 ) = 600 so we know that there are 60 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 600 = 610 20 + 590 = 610 30 + 580 = 610 etc . since there are 60 total terms , this pattern will create 30 ' pairs ' of 610 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 30 ) ( 610 ) / ( 60 ) = 305 final answer : c"
a ) 190 , b ) 195 , c ) 305 , d ) 205 , e ) 210
c
divide(divide(multiply(add(10, 600), add(divide(subtract(600, 10), 10), const_1)), const_2), add(divide(subtract(600, 10), 10), const_1))
add(n0,n2)|subtract(n2,n0)|divide(#1,n0)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)|
general
a $ 74.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ?
"listed selling price of chair = 74.95 $ discounted selling price of chair = 59.95 $ discount = 74.95 - 59.95 = 15 $ % decrease in price of chair = ( 15 / 74.95 ) * 100 % = 20 % approx answer b"
a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 80 %
b
multiply(divide(subtract(74.95, 59.95), 74.95), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
general
if rs . 595 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ?
"( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 595 * 2 / 17 = rs . 70 b ' s share = 595 * 3 / 17 = rs . 105 c ' s share = 595 * 12 / 17 = rs . 420 . answer : a"
a ) s . 420 , b ) s . 360 , c ) s . 389 , d ) s . 368 , e ) s . 323
a
divide(595, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1))
divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)|
general
a part - time employee ’ s hourly wage was increased by 50 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ?
let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 50 % wage increase , the employee makes $ 1.50 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.50 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.50 to get : new # of hours = 100 / 1.50 β‰ˆ 67 hours so , the number of hours decreases from 100 hours to ( approximately ) 67 hours . this represents a 33 % decrease ( approximately ) . answer : d
a ) 9 % , b ) 15 % , c ) 25 % , d ) 33 % , e ) 50 %
d
multiply(divide(divide(50, const_100), divide(add(50, const_100), const_100)), const_100)
add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)
general
cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 3 % sugar by weight . to make a delicious and healthy mixture that is 5 % sugar , what should be the ratio of cereal a to cereal b , by weight ?
"ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 5 - 3 ) / ( 10 - 5 ) = 2 / 5 so they should be mixed in the ratio 2 : 5 answer - a"
a ) 2 : 5 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3
a
divide(subtract(5, 3), subtract(10, 5))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|
general
the q train leaves station a moving at a constant speed , and passes by stations b and c , in this order . it takes the q train 7 hours to reach station b , and 5 additional hours to reach station c . the distance between stations a and b is m kilometers longer than the distance between stations b and c . what is the distance between stations a and c in terms of m ?
the reason it is failing for you is that you chose incorrect numbers . if the question says it q took 7 hrs to reach from a to b and 5 hrs to reach from b to c at a constant speed . it shows that distance ab and bc should be in ratio of 7 / 5 . if you take such numbers you can solve problem . ab = 7 , bc = 5 therefore ab - bc = 2 but from question , ab - bc = m = > m = 2 now total distance = ab + bc = 12 substitute 12 to get answer in terms of m total distance = 12 = 6 m ans b
a ) 1.8 m , b ) 6 m , c ) 7 m , d ) 9 m , e ) 12 m
b
subtract(divide(const_2, subtract(const_1, divide(5, 7))), const_1)
divide(n1,n0)|subtract(const_1,#0)|divide(const_2,#1)|subtract(#2,const_1)
physics
what is the area of a square field whose sides have a length of 12 meters ?
"12 * 12 = 144 sq m the answer is c ."
a ) 225 sq m , b ) 126 sq m , c ) 144 sq m , d ) 267 sq m , e ) 231 sq m
c
divide(square_area(12), const_2)
square_area(n0)|divide(#0,const_2)|
geometry
if 4 a – 2 b – 2 c = 32 and √ 3 a - √ ( 2 b + 2 c ) = 4 , what is the value of a + b + c ?
"when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a – 2 b – 2 c = 32 can be written as √ 3 a ^ 2 - √ ( 2 b + 2 c ) ^ 2 = 32 { √ 3 a - √ ( 2 b + 2 c ) } { √ 3 a + √ ( 2 b + 2 c ) } = 32 . . or 4 * √ 3 a + √ ( 2 b + 2 c ) = 32 . . or √ 3 a + √ ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) √ 3 a - √ ( 2 b + 2 c ) = 4 . . 2 ) √ 3 a + √ ( 2 b + 2 c ) = 8 . . a ) add the two eq . . √ 3 a + √ ( 2 b + 2 c ) + √ 3 a - √ ( 2 b + 2 c ) = 12 . . 2 √ 3 a = 12 . . or √ 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . √ 3 a + √ ( 2 b + 2 c ) - √ 3 a + √ ( 2 b + 2 c ) = 4 . . 2 √ ( 2 b + 2 c ) = 4 . . √ ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . d"
a ) 3 , b ) 9 , c ) 10 , d ) 12 , e ) 14
d
subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(32, 4), 4), 2), 2), const_1), 4), const_1), 2), 4), 4)
subtract(n3,n7)|subtract(#0,n7)|subtract(#1,n2)|subtract(#2,n2)|subtract(#3,const_1)|subtract(#4,n0)|subtract(#5,const_1)|add(n2,#6)|subtract(#7,n0)|subtract(#8,n7)|
general
a room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room .
"let us calculate both the length and width of the room in centimeters . length = 6 meters and 24 centimeters = 624 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 624 and 432 = 48 thus , the number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 answer : b"
a ) 107 , b ) 117 , c ) 127 , d ) 137 , e ) 147
b
divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32))))
multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)|
general
how many integers are between 8 and 122 / 7 , inclusive ?
122 / 7 = 17 . xx we are not concerned about the exact value of 122 / 7 as we just need the integers . the different integers between 8 and 122 / 7 would be 8 , 9 , 10 , 11 , 12,13 , 14,15 , 16,17 total number of integers = 10 option b
a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 16
b
add(subtract(divide(122, 7), 8), const_1)
divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)
general
what percent of 10 is 10 percent of 1 ?
"10 percent of 1 is equal to 0.1 so we have to compute what percent of 10 is 0.1 . so this means we have to compute what percent of 100 is 1 . . . . so 1 % answer b ."
a ) 0.1 % , b ) 1 % , c ) 10 % , d ) 90 % , e ) 100 %
b
p_after_gain(10, 10)
p_after_gain(n1,n0)|
gain
there are 60 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 14 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ?
"x + ( x - 14 ) = 60 - - > x = 37 . answer : b ."
a ) 20 , b ) 37 , c ) 42 , d ) 53 , e ) 64
b
divide(add(60, 14), const_2)
add(n0,n1)|divide(#0,const_2)|
general
a man sitting in a train which is traveling at 50 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ?
relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 50 ) kmph = 62 kmph . answer : c .
a ) 50 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph
c
subtract(multiply(divide(280, 9), const_3_6), 50)
divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)
physics
the cricket team of 11 members is 26 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of the remainingplayers is 1 year less than the average age of the whole team . what is the average age of the team ?
"let the average age of the whole team be x years . 11 x - ( 26 + 29 ) = 9 ( x - 1 ) = > 11 x - 9 x = 46 = > 2 x = 46 = > x = 23 . so , average age of the team is 23 years . c"
a ) 18 , b ) 21 , c ) 23 , d ) 25 , e ) 27
c
divide(subtract(add(add(26, 3), 26), subtract(11, 2)), subtract(11, subtract(11, 2)))
add(n1,n2)|subtract(n0,n3)|add(n1,#0)|subtract(n0,#1)|subtract(#2,#1)|divide(#4,#3)|
general
the true discount on a bill of rs . 2160 is rs . 360 . what is the banker ' s discount ?
"explanation : f = rs . 2160 td = rs . 360 pw = f - td = 2160 - 360 = rs . 1800 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 1800 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2160 for unexpired time = 360 / 1800 Γ— 2160 = 15 Γ— 2160 = rs . 432 answer : option a"
a ) rs . 432 , b ) rs . 422 , c ) rs . 412 , d ) rs . 442 , e ) none of these
a
multiply(divide(360, subtract(2160, 360)), 2160)
subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)|
gain
if the cost price is 82 % of the selling price , then what is the profit percent ?
"let s . p . = $ 100 c . p . = $ 82 profit = $ 18 profit % = 18 / 82 * 100 = 25 / 6 = 22 % approximately answer is b"
a ) 5 % , b ) 22 % , c ) 13 % , d ) 21 % , e ) 19 %
b
multiply(divide(subtract(const_100, 82), 82), const_100)
subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
the average of first six prime numbers which are between 50 and 90 is
"explanation : first six prime numbers which are between 50 and 90 = 53 , 59 , 61 , 67 , 71 , 73 average = ( 53 + 59 + 61 + 67 + 71 + 73 ) / 6 = 64 answer : d"
a ) 35.4 , b ) 42 , c ) 45.7 , d ) 64 , e ) 67
d
add(50, const_1)
add(n0,const_1)|
general
if 0.5 % of a = 85 paise , then the value of a is ?
"answer ∡ 0.5 / 100 of a = 85 / 100 ∴ a = rs . ( 85 / 0.5 ) = rs . 170 correct option : a"
a ) rs . 170 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none
a
divide(85, 0.5)
divide(n1,n0)|
gain
suganya and suriya are partners in a business . suganya invests rs . 15,000 for 8 months and suriya invests rs . 42,000 for 10 months . out of a profit of rs . 31,570 . suganya ' s share is
"solution ratio of their shares = ( 15000 Γ£ β€” 8 ) : ( 42000 Γ£ β€” 10 ) = 2 : 7 . suganya ' s share = rs . ( 31570 Γ£ β€” 2 / 9 ) = rs . 7015.56 answer d"
a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 18,040 , d ) rs . 7015.56 , e ) none
d
multiply(multiply(const_0_25, const_100), 10)
multiply(const_0_25,const_100)|multiply(n3,#0)|
gain
find the compound interest on $ 1200 for 5 years at 20 % p . a . if ci is component yearly ?
"a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 5 = $ 2986 ci = $ 1786 answer is c"
a ) $ 120 , b ) $ 150 , c ) $ 1786 , d ) $ 250 , e ) $ 300
c
subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 5)), 1200)
divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)|
gain
a person want to give his money of $ 15800 to his 4 children a , b , c , d in the ratio 5 : 9 : 6 : 5 . what is the a + c share ?
"a ' s share = 15800 * 5 / 25 = $ 3160 c ' s share = 15800 * 6 / 25 = $ 3792 a + c = $ 6952 answer is d"
a ) $ 4000 , b ) $ 2890 , c ) $ 3005 , d ) $ 6952 , e ) $ 6003
d
multiply(divide(const_3.0, add(add(5, const_3.0), 6)), 15800)
add(const_3.0,const_2.0)|add(const_4.0,#0)|divide(n1,#1)|multiply(n0,#2)|
other
a certain class of students is being divided into teams . the class can either be divided into 12 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?
"let total no of students in the class be n so , we are told that n is divisible by both 12 24 so , lets find the least common multiple of 12 24 , ie 24 so our answer is ( a ) 24"
a ) 24 , b ) 36 , c ) 48 , d ) 60 , e ) 72
a
lcm(12, 24)
lcm(n0,n1)|
general
a group of 6 children and 10 adults are going to the zoo . child tickets cost $ 10 , and adult tickets cost $ 16 . how much will the zoo tickets cost in all ?
step 1 : find the cost of the children ' s tickets . 6 Γ— $ 10 = $ 60 step 2 : find the cost of the adults ' tickets . 10 Γ— $ 16 = $ 160 step 3 : find the total cost . $ 60 + $ 160 = $ 220 the zoo tickets will cost $ 220 . answer is a .
a ) $ 220 , b ) $ 340 , c ) $ 150 , d ) $ 100 , e ) $ 120
a
add(multiply(6, 10), multiply(10, 16))
multiply(n0,n1)|multiply(n1,n3)|add(#0,#1)
general
the price of rice falls by 40 % . how much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously ?
"solution : let rs . 100 be spend on rice initially for 20 kg . as the price falls by 40 % , new price for 20 kg rice , = ( 100 - 40 % of 100 ) = 60 new price of rice = 60 / 20 = rs . 3 per kg . rice can bought now at = 100 / 3 = 33.33 kg . answer : option e"
a ) 5 kg , b ) 15 kg , c ) 25 kg , d ) 30 kg , e ) none
e
divide(const_100, divide(subtract(const_100, 40), 20))
subtract(const_100,n0)|divide(#0,n1)|divide(const_100,#1)|
gain
the sum of the squares of three numbers is 138 , while the sum of their products taken two at a time is 131 . find the sum ?
"a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 a + b + c ^ 2 = 400 root ( 400 ) = 20 answer e"
a ) 30 , b ) 15 , c ) 40 , d ) 45 , e ) 20
e
add(multiply(sqrt(divide(subtract(138, 131), const_2)), const_100), sqrt(subtract(138, divide(subtract(138, 131), const_2))))
subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)|
general
divide rs . 1200 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ?
"a + b + c = 1200 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1200 = > 300 answer : e"
a ) s . 800 , b ) s . 400 , c ) s . 600 , d ) s . 500 , e ) s . 300
e
divide(1200, const_3)
divide(n0,const_3)|
general
108 . triangle a ’ s base is 9 % greater than the base of triangle b , and a ’ s height is 9 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ?
"wish the question specified that we are talking about corresponding height . base of a = 10 / 9 * base of b height of a = 8 / 9 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 10 / 9 * 8 / 9 * area of b = 80 / 81 * area of b area of a is 1.3 % more than the area of b . answer ( d )"
a ) 9 % less , b ) 1 % less , c ) equal to each other , d ) 1.3 % more , e ) 9 % more
d
divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 9), subtract(const_100, 9))))
add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4)|
geometry
tourist purchased a total of 30 travelers checks in $ 50 and $ 100 denominations . the total worth of the travelers checks is $ 1800 . how many checks of $ 50 denominations can he spend so that average amount ( arithmetic mean ) of the remaining travelers checks is $ 62.5 ?
"you could set - up a quick table and brute force the answer . a 6 * 50 300 1800 - 300 1500 24 62.50 b 16 * 50 600 1800 - 600 1200 18 66.67 c 15 * 50 750 1800 - 750 1050 15 70.00 d 20 * 50 1000 1800 - 1000 800 10 80.00 e 24 * 50 1200 1800 - 1200 600 6 100.00 answer is a"
a ) 6 , b ) 12 , c ) 15 , d ) 20 , e ) 24
a
divide(subtract(multiply(62.5, 30), 1800), subtract(62.5, 50))
multiply(n0,n5)|subtract(n5,n1)|subtract(#0,n3)|divide(#2,#1)|
general
what will be the compound interest on rs . 40000 after 3 years at the rate of 12 % per annum
"explanation : ( 40000 Γ— ( 1 + 12 / 100 ) 3 ) = > 40000 Γ— 28 / 25 Γ— 28 / 25 Γ— 28 / 25 = > 56197.12 so compound interest will be 56197.12 - 40000 = rs 16197.12 option c"
a ) rs 16123.20 , b ) rs 16123.30 , c ) rs 16197.12 , d ) rs 16123.50 , e ) none of these
c
subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100))
divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)|
gain
how many pieces of 75 cm can be cut from a rope 54 meters long ?
"explanation : total pieces of 75 cm that can be cut from a rope of 54 meters long is = ( 54 meters ) / ( 75 cm ) = ( 54 meters ) / ( 0.75 meters ) = 72 answer : d"
a ) 30 , b ) 40 , c ) 60 , d ) 72 , e ) can not be determined
d
divide(54, 75)
divide(n1,n0)|
physics
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 16 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ?
"the time it takes printer x is 16 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 16 / ( 20 / 3 ) = 3 * 16 / 20 = 12 / 5 the answer is c ."
a ) 7 / 4 , b ) 10 / 3 , c ) 12 / 5 , d ) 15 / 7 , e ) 18 / 11
c
divide(16, divide(const_1, add(divide(const_1, 10), divide(const_1, 20))))
divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)|
general