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9 log 9 ( 6 ) = ?
"exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 6 ) = 6 correct answer b"
a ) 1 , b ) 6 , c ) 3 , d ) 4 , e ) 5
b
divide(log(multiply(9, 9)), log(const_10))
log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)|
other
two mixtures of a and b have a and b in the ratio 3 : 2 and 3 : 4 . in what proportion should these two mixtures be mixed to get a new mixture in which the ration of a to b is 5 : 4 ?
i got c but with a different , yet slightly longer approach , which i constantly use for proportion problems such as this one . usually , this approach is very efficient when used with slightly difficult questions , but this problem is quite hard so i had to slightly tweak it . i will show you how i usually use the approach and later show you how to use it with this specific problem . example . 1 liter of solution a contains 45 % alcohol , while 1 liter of solution b contains 20 % alcohol . in what ratio must the two solutions be used to get a solution with 30 % alcohol solution : 1 . 45 / 100 * [ a / ( a + b ) ] + 20 / 100 * [ b / ( a + b ) ] = 30 / 100 2 . multiply 100 to both sides to arrive at 45 a / ( a + b ) + 20 b / ( a + b ) = 30 3 . multiply ( a + b ) to both sides to arrive at 45 a + 20 b = 30 a + 30 b 4 . distribute to arrive at 15 a = 10 b 5 . thus the ratio is a / b = 10 / 15 = 2 / 3 now using this same approach , we tackle gopu 106 ’ s question . it is important to first think of x in the mixture as the alcohol in the problem above ; hence , a mixture of x and y in the ratio of 3 : 2 translates to x is 3 / 5 of the solution . applying this concept to all three equations , we write : 1 . 3 / 5 * [ a / ( a + b ) ] + 3 / 7 * [ b / ( a + b ) ] = 5 / 9 2 . now here is the tweak that must be made to continue with this approach . you must find the common denominator for all three numbers and organize the fractions accordingly . by finding the common denominator of 5 , 79 ( or 315 ) we re - write the equations as follows 3 . 189 / 315 * [ a / ( a + b ) ] + 135 / 315 * [ b / ( a + b ) ] = 175 / 315 4 . multiply 315 to both sides to arrive at 189 a / ( a + b ) + 135 b / ( a + b ) = 175 5 . multiply ( a + b ) to both sides to arrive at 189 a + 135 b = 175 a + 175 b 6 . distribute to arrive at 14 a = 40 b 7 . thus the ratio is a / b = 40 / 14 = 20 / 7 or answer c
a ) 6 : 1 , b ) 5 : 4 , c ) 20 : 7 , d ) 10 : 9 , e ) 14 : 11
c
divide(subtract(divide(5, add(5, 4)), divide(const_3, add(3, 4))), subtract(divide(3, 5), divide(5, add(5, 4))))
add(n3,n4)|add(n0,n3)|divide(n0,n4)|divide(n4,#0)|divide(const_3,#1)|subtract(#3,#4)|subtract(#2,#3)|divide(#5,#6)
other
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 60 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ?
"this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 60 % is one st . dev , then on either side of m we have 60 / 2 = 30 % . so , 30 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 30 = 80 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 80 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it a"
a ) 80 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
a
subtract(const_100, divide(subtract(const_100, 60), const_2))
subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)|
general
8 kilograms of rice costing rs . 16 per kg is mixed with 4 kilograms of rice costing rs . 22 per kg . what is the average price of the mixture ?
p 1 = rs . 16 per kg , p 2 = rs . 22 per kg , q 1 = 8 kg , q 2 = 4 kg now , p = ( p 1 q 1 + p 2 q 2 ) / ( q 1 + q 2 ) average price of the mixture = 8 * 16 + 4 * 22 / 12 = 128 + 88 / 12 = 216 / 12 = 18 answer : b
a ) 20 , b ) 18 , c ) 16 , d ) 19 , e ) 17
b
divide(add(multiply(22, 4), multiply(8, 16)), add(8, 4))
add(n0,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#1,#2)|divide(#3,#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 70 metres ?
"speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 70 / 5 = 14 seconds . answer : e"
a ) 22 seconds , b ) 65 seconds , c ) 78 seconds , d ) 12 seconds , e ) 14 seconds
e
divide(70, multiply(add(15, 3), const_0_2778))
add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
if the numbers 1 to 99 are written on 99 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ?
"there are 25 primes , 73 composite numbers from 1 to 100 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 99 . answer : c"
a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 99 , d ) 1 , e ) 2
c
divide(1, 99)
divide(n0,n1)|
other
a trader purchased two colour televisions for a total of rs . 35000 . he sold one colour television at 30 % profit and the other 40 % profit . find the difference in the cost prices of the two televisions if he made an overall profit of 32 % ?
let the cost prices of the colour television sold at 30 % profit and 40 % profit be rs . x and rs . ( 35000 - x ) respectively . total selling price of televisions = x + 30 / 100 x + ( 35000 - x ) + 40 / 100 ( 35000 - x ) = > 130 / 100 x + 140 / 100 ( 35000 - x ) = 35000 + 32 / 100 ( 35000 ) x = 28000 35000 - x = 7000 difference in the cost prices of televisions = rs . 21000 answer : a
a ) 21000 , b ) 21009 , c ) 21029 , d ) 21298 , e ) 21098
a
subtract(subtract(35000, divide(subtract(multiply(divide(add(const_100, 32), const_100), 35000), multiply(divide(add(const_100, 30), const_100), 35000)), subtract(divide(add(const_100, 40), const_100), divide(add(const_100, 30), const_100)))), divide(subtract(multiply(divide(add(const_100, 32), const_100), 35000), multiply(divide(add(const_100, 30), const_100), 35000)), subtract(divide(add(const_100, 40), const_100), divide(add(const_100, 30), const_100))))
add(n3,const_100)|add(n1,const_100)|add(n2,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|multiply(n0,#3)|multiply(n0,#4)|subtract(#5,#4)|subtract(#6,#7)|divide(#9,#8)|subtract(n0,#10)|subtract(#11,#10)
gain
an article is bought for rs . 775 and sold for rs . 900 , find the gain percent ?
"775 - - - - 125 100 - - - - ? = > 16.1 % answer : e"
a ) 11.1 % , b ) 12.1 % , c ) 13.1 % , d ) 15.1 % , e ) 16.1 %
e
subtract(const_100, divide(multiply(900, const_100), 775))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|
gain
9 ! / ( 9 - 3 ) ! = ?
9 ! / ( 9 - 3 ) ! = 9 ! / 6 ! = 9 * 8 * 7 = 504 . hence , the correct answer is e .
a ) 336 , b ) 346 , c ) 356 , d ) 366 , e ) 504
e
divide(factorial(9), factorial(subtract(9, 3)))
factorial(n0)|subtract(n0,n2)|factorial(#1)|divide(#0,#2)
general
if 213 × 16 = 3408 , then 1.6 × 21.3 is equal to :
"solution 1.6 × 21.3 = ( 16 / 10 x 213 / 10 ) = ( 16 x 213 / 100 ) = 3408 / 100 = 34.08 . answer c"
a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these
c
multiply(1.6, 21.3)
multiply(n4,n3)|
general
if the simple interest on a sum of money for 5 years at 18 % per annum is rs . 900 , what is the compound interest on the same sum at the rate and for the same time ?
"sum = ( 900 * 100 ) / ( 5 * 18 ) = rs . 1 , 000.00 c . i . on rs . rs . 1 , 000.00 for 5 years at 18 % = rs . 2 , 287.76 . = rs . 2 , 287.76 - 1 , 000.00 = rs . 1287.76 answer : a"
a ) rs . 1287.76 , b ) rs . 1284.76 , c ) rs . 1587.76 , d ) rs . 1266.76 , e ) rs . 1283.76
a
subtract(add(add(divide(multiply(divide(900, multiply(divide(18, const_100), 5)), 18), const_100), divide(900, multiply(divide(18, const_100), 5))), divide(multiply(add(divide(multiply(divide(900, multiply(divide(18, const_100), 5)), 18), const_100), divide(900, multiply(divide(18, const_100), 5))), 18), const_100)), divide(900, multiply(divide(18, const_100), 5)))
divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|
gain
the true discount on a bill due 9 months hence at 16 % per annum is rs . 150 . the amount of the bill is
explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 150 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 150 or x = 1250 . p . w . = rs . 1250 sum due = p . w . + t . d . = rs . ( 1250 150 ) = rs . 1400 answer : c
a ) 1200 , b ) 1764 , c ) 1400 , d ) 1354 , e ) none of these
c
add(divide(150, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 150)
multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)
gain
andrew travelling to 7 cities . gasoline prices varied from city to city . $ 1.75 , $ 1.61 , $ 1.79 , $ 2.11 , $ 1.96 , $ 2.09 , $ 1.81 . what is the median gasoline price ?
"ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.81 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.81 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b"
a ) $ 1 , b ) $ 1.81 , c ) $ 1.92 , d ) $ 2.13 , e ) $ 2.15
b
min(divide(add(add(add(add(add(add(1.75, 1.61), 1.79), 2.11), 1.96), 2.09), 1.81), 7), 1.81)
add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|add(n6,#3)|add(n7,#4)|divide(#5,n0)|min(n7,#6)|
general
a freight elevator can carry a maximum load of 1100 pounds . sean , who weighs 200 pounds , is in the elevator with two packages weighing 150 pounds and 280 pounds . if he needs to fit 3 more packages in the elevator that weigh as much as possible without exceeding the elevator limit , what is the difference between their average and the average of the two packages already in the elevator ?
the average of existing 2 package is 150 + 280 / 2 = 430 / 2 = 215 remaining allowed weight = 1100 - 200 - 430 = 470 . allowed per package = 470 / 3 = 156 so difference in average of existing and allowable = 215 - 156 = 59 hence a
a ) 59 , b ) 85 , c ) 190 , d ) 215 , e ) 210
a
subtract(divide(add(150, 280), const_2), divide(subtract(subtract(1100, 200), add(150, 280)), 3))
add(n2,n3)|subtract(n0,n1)|divide(#0,const_2)|subtract(#1,#0)|divide(#3,n4)|subtract(#2,#4)
general
power windows : 60 % anti - lock brakes : 25 % cd player : 75 % the table above shows the number of vehicles at bill ' s car dealership that have certain features . no vehicle has all 3 features , but 10 % have power windows and anti - lock brakes , 15 % have anti - lock brakes and a cd player , and 22 % have power windows and a cd player . what percent of the vehicles at bill ' s car dealership have a cd player but no power windows or anti - lock brakes ?
answer : d were looking for the number of cars with a cd player but no other features . we know that 40 % of the cars have a cd player , 15 % have a cd player and anti - lock brakes , while 22 % have a cd player and power windows . since no car has all three features , those account for all of the possibilities except for what were looking for . if a car has a cd player , it must either have anti - lock brakes , power windows , or no other features . since the total of cars with a cd player is 75 % , we can set up the following equation : 75 = 15 + 22 + x x = 38 , choiced .
a ) 25 , b ) 18 , c ) 11 , d ) 38 , e ) 0
d
subtract(subtract(75, 15), 22)
subtract(n2,n5)|subtract(#0,n6)
gain
a train 360 m long is running at a speed of 60 km / hr . in what time will it pass a bridge 140 m long ?
"speed = 60 * 5 / 18 = 50 / 3 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 3 / 50 = 30 sec answer : b"
a ) 40 sec , b ) 30 sec , c ) 26 sec , d ) 27 sec , e ) 34 sec
b
divide(360, multiply(subtract(60, 140), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
the value of ( log 9 27 + log 8 32 ) is :
let log 9 27 = x . then , 9 x = 27 = > ( 32 ) x = 33 = > 2 x = 3 = > x = 3 / 2 let log 8 32 = y . then 8 y = 32 = > ( 23 ) y = 25 = > 3 y = 5 = > y = 5 / 3 answer : c
a ) 7 / 2 , b ) 19 / 6 , c ) 5 / 3 , d ) 7 , e ) 8
c
divide(log(32), log(8))
log(n3)|log(n2)|divide(#0,#1)
other
a meal cost $ 37.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be :
"10 % ( 37.5 ) = 3.75 15 % ( 37.5 ) = 5.625 total amount could have been 37.5 + 3.75 and 37.5 + 5.625 = > could have been between 41.25 and 43.125 = > approximately between 41 and 43 answer is a ."
a ) 41 - 43 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37
a
add(multiply(37.50, divide(15, const_100)), 37.50)
divide(n2,const_100)|multiply(n0,#0)|add(n0,#1)|
general
find the slope of the line perpendicular to the line y = ( 1 / 4 ) x - 7
"two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 4 . if m is the slope of the line perpendicular to the given line , then m × ( 1 / 4 ) = - 1 solve for m m = - 4 correct answer c ) - 4"
a ) 1 , b ) 2 , c ) - 4 , d ) 4 , e ) 5
c
divide(1, 4)
divide(n0,n1)|
general
a person can row at 9 kmph and still water . he takes 7 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ?
"let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 7.5 = > x / 10 + x / 8 = 15 / 2 = > ( 4 x + 5 x ) / 40 = 15 / 2 = > x = 33 km . answer : c"
a ) 60 km , b ) 87 km , c ) 33 km , d ) 67 km , e ) 20 km
c
divide(multiply(multiply(subtract(9, 1), add(9, 1)), 7), add(add(9, 1), subtract(9, 1)))
add(n0,n2)|subtract(n0,n2)|add(#0,#1)|multiply(#0,#1)|multiply(n1,#3)|divide(#4,#2)|
physics
walking with 4 / 5 of my usual speed , i miss the bus by 4 minutes . what is my usual time ?
"speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 5 1 - - - - - - - - 4 4 - - - - - - - - - ? è 16 answer : a"
a ) 16 min , b ) 26 min , c ) 34 min , d ) 20 min , e ) 12 min
a
multiply(divide(4, divide(5, 4)), 5)
divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)|
physics
find the principal which yields a simple interest of rs . 20 and compound interest of rs . 22 in two years , at the same percent rate per annum ?
explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 22 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 22 – 20 ) / 20 ] * 100 = 10 % p . a . let the principal be rs . x time = t = 2 years % rate = 10 % p . a . si = ( prt / 100 ) 20 = ( x * 10 * 2 ) / 100 x = rs . 100 answer : d
a ) s . 520 , b ) s . 480 , c ) s . 420 , d ) s . 100 , e ) s . 200
d
divide(multiply(subtract(22, 20), const_100), const_2)
subtract(n1,n0)|multiply(#0,const_100)|divide(#1,const_2)
gain
if x / ( 11 p ) is an odd prime number , where x is a positive integer and p is a prime number , what is the least value of x ?
x / ( 11 p ) = odd prime number x = odd prime number * 11 p least value of x = lowest odd prime number * 11 * lowest value of p = 3 * 11 * 2 = 66 answer d
a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99
d
multiply(multiply(const_3, 11), const_2)
multiply(n0,const_3)|multiply(#0,const_2)
general
9 persons went to a hotel for taking their meals . 8 of them spent rs . 12 each on their meals and the ninth spent rs . 8 more than the average expenditure of all the 9 . what was the total money spent by them .
solution : let the average expenditure of all the nine be x . then , 12 × 8 + ( x + 8 ) = 9 x . therefore , x = 13 . total money spent , = 9 x = rs . ( 9 × 13 ) = rs . 117 answer : option c
a ) rs . 115 , b ) rs . 116 , c ) rs . 117 , d ) rs . 118 , e ) none
c
add(add(12, 9), multiply(8, 12))
add(n0,n2)|multiply(n1,n2)|add(#0,#1)
general
a cube of edge 6 cm is cut into cubes each of edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to :
"sol . required ratio = 6 * 1 * 1 / 6 * 6 * 6 = 1 / 36 = 1 : 36 . answer b"
a ) 1 : 25 , b ) 1 : 36 , c ) 1 : 52 , d ) 1 : 522 , e ) none
b
divide(const_4, const_100)
divide(const_4,const_100)|
geometry
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 36 . what is the difference between the two digits of that number ?
"sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 36 ⇔ 9 ( x - y ) = 36 ⇔ x - y = 4 answer a"
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9
a
divide(36, subtract(const_10, const_1))
subtract(const_10,const_1)|divide(n0,#0)|
general
a snail , climbing a 66 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ?
"total transaction in two days = 4 - 2 = 2 feet in 62 days it will climb 62 feet on the 63 rd day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 63 b"
a ) 42 , b ) 63 , c ) 77 , d ) 80 , e ) 91
b
subtract(66, 4)
subtract(n0,n1)|
physics
if 6 pr = 360 and if 6 cr = 15 , find r ?
solution : npr = ncr � r ! 6 pr = 15 * r ! 360 = 15 * r ! r ! = 360 / 15 = 24 r ! = 4 * 3 * 2 * 1 = > r ! = 4 ! therefore , r = 4 . answer : option c
a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 2
c
divide(divide(360, 15), 6)
divide(n1,n3)|divide(#0,n0)
general
what is the sum of all remainders obtained when the first 130 natural numbers are divided by 9 ?
"a positive integer can give only the following 9 remainders when divided by 9 : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 0 . 1 divided by 9 gives the remainder of 1 ; 2 divided by 9 gives the remainder of 2 ; . . . 8 divided by 9 gives the remainder of 8 ; 9 divided by 9 gives the remainder of 0 . we ' ll have 11 such blocks , since 99 / 9 = 11 . the last will be : 91 divided by 9 gives the remainder of 1 ; 92 divided by 9 gives the remainder of 2 ; . . . 98 divided by 9 gives the remainder of 8 ; 99 divided by 9 gives the remainder of 0 . the last number , 100 , gives the remainder of 1 when divided by 9 , thus the sum of all remainders will be : 11 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0 ) + 1 = 412 . answer : d ."
a ) 397 , b ) 401 , c ) 403 , d ) 412 , e ) 399
d
add(const_3, multiply(divide(subtract(130, reminder(130, 9)), 9), add(add(add(const_1, const_2), const_3), add(add(const_1, const_4), add(const_4, add(add(const_1, const_2), const_3))))))
add(const_1,const_2)|add(const_1,const_4)|reminder(n0,n1)|add(#0,const_3)|subtract(n0,#2)|add(#3,const_4)|divide(#4,n1)|add(#1,#5)|add(#3,#7)|multiply(#8,#6)|add(#9,const_3)|
general
by selling 90 pens , a trader gains the cost of 15 pens . find his gain percentage ?
"let the cp of each pen be rs . 1 . cp of 90 pens = rs . 90 profit = cost of 15 pens = rs . 15 profit % = 15 / 90 * 100 = 16.66 % answer : a"
a ) 16.66 % , b ) 17.66 % , c ) 18.66 % , d ) 19.66 % , e ) 20.66 %
a
multiply(divide(15, 90), const_100)
divide(n1,n0)|multiply(#0,const_100)|
gain
how many paying stones , each measuring 2 m * 2 m are required to pave a rectangular court yard 30 m long and 18 m board ?
"30 * 18 = 2 * 2 * x = > x = 135 answer : c"
a ) 99 , b ) 18 , c ) 135 , d ) 17 , e ) 12
c
divide(rectangle_area(30, 18), rectangle_area(2, 2))
rectangle_area(n2,n3)|rectangle_area(n0,n1)|divide(#0,#1)|
physics
a jogger running at 9 km / hr along side a railway track is 250 m ahead of the engine of a 100 m long train running at 99 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 99 - 9 = 90 km / hr . = 90 * 5 / 18 = 25 m / sec . distance to be covered = 250 + 100 = 350 m . time taken = 350 / 25 = 14 sec . answer : a"
a ) 14 sec , b ) 67 sec , c ) 98 sec , d ) 36 sec , e ) 23 sec
a
divide(add(250, 100), multiply(subtract(99, 9), divide(divide(const_10, const_2), divide(subtract(99, 9), const_2))))
add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|
general
a train 200 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is ?
"speed of the train relative to man = ( 200 / 20 ) m / sec = 5 m / sec . [ 5 * ( 18 / 5 ) ] km / hr = 18 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 18 = = > x = 23 km / hr . answer : e"
a ) 36 , b ) 50 , c ) 88 , d ) 66 , e ) 23
e
divide(divide(subtract(200, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778)
multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
a pupil ' s marks were wrongly entered as 73 instead of 65 . due to the average marks for the class got increased by half . the number of pupils in the class is ?
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 65 ) = > x / 2 = 8 = > x = 16 . answer : c"
a ) 18 , b ) 82 , c ) 16 , d ) 27 , e ) 29
c
multiply(subtract(73, 65), const_2)
subtract(n0,n1)|multiply(#0,const_2)|
general
the average salary of 16 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ?
total salary . . . 16 * 20 k = 320 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 7 emp sal = 320 k - 125 k - 64 k = 131 k average = 131 k / 7 = 18714 ans : a
a ) $ 18,714 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 12,300
a
divide(subtract(subtract(multiply(multiply(5, 4), multiply(const_4, const_4)), multiply(multiply(5, 5), 5)), multiply(4, 16)), add(const_2, 5))
add(n2,const_2)|multiply(n2,n4)|multiply(const_4,const_4)|multiply(n2,n2)|multiply(n0,n4)|multiply(#1,#2)|multiply(n2,#3)|subtract(#5,#6)|subtract(#7,#4)|divide(#8,#0)
general
a shopkeeper sold an article at $ 100 with 30 % profit . then find its cost price ?
"cost price = selling price * 100 / ( 100 + profit ) c . p . = 100 * 100 / 130 = $ 77 ( approximately ) answer is d"
a ) $ 120 , b ) $ 100 , c ) $ 91 , d ) $ 77 , e ) $ 69
d
multiply(100, divide(100, add(100, 30)))
add(n1,n0)|divide(n0,#0)|multiply(n0,#1)|
gain
in an election between two candidates , the first candidate got 80 % of the votes and the second candidate got 480 votes . what was the total number of votes ?
"let v be the total number of votes . 0.2 v = 480 v = 2400 the answer is e ."
a ) 1600 , b ) 1800 , c ) 2000 , d ) 2200 , e ) 2400
e
divide(multiply(480, const_100), subtract(const_100, 80))
multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|
gain
peter invested a certain sum of money in a simple interest bond whose value grew to $ 400 at the end of 4 years and to $ 500 at the end of another 2 years . what was the rate of interest in which he invested his sum ?
"lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of 4 years the principal p amounts to $ 400 and after a time of 6 years ( question says after another 6 years so 4 + 2 ) p becomes $ 500 . formulating the above data amount ( a 1 ) at end of 4 years a 1 = p ( 1 + 4 r / 100 ) = 400 amount ( a 2 ) at end of 6 years a 2 = p ( 1 + 2 r / 100 ) = 500 dividing a 2 by a 1 we get ( 1 + 4 r / 100 ) / ( 1 + 2 r / 100 ) = 5 / 4 after cross multiplication we are left with 4 r = 100 which gives r = 25 % option : d"
a ) 12 % , b ) 12.5 % , c ) 67 % , d ) 25 % , e ) 33 %
d
multiply(divide(divide(subtract(500, 400), 2), subtract(400, multiply(divide(subtract(500, 400), 2), 4))), const_100)
subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)|
gain
twenty - one dots are evenly spaced on the circumference of a circle . how many combinations of three dots can we make from these 21 dots that do not form an equilateral triangle ?
the total number of ways we can choose three dots is 21 c 3 = 1330 . we can form seven equilateral triangles from these twenty - one dots . there are 1330 - 7 = 1323 combinations which do not form an equilateral triangle . the answer is e .
['a ) 1299', 'b ) 1305', 'c ) 1311', 'd ) 1317', 'e ) 1323']
e
subtract(divide(divide(multiply(multiply(21, subtract(21, const_1)), subtract(subtract(21, const_1), const_1)), const_3), const_2), add(const_3, const_4))
add(const_3,const_4)|subtract(n0,const_1)|multiply(n0,#1)|subtract(#1,const_1)|multiply(#2,#3)|divide(#4,const_3)|divide(#5,const_2)|subtract(#6,#0)
geometry
if the area of circle is 616 sq cm then its circumference ?
the area of circle = pie * r ^ 2 = 616 = > r = 14 2 * 22 / 7 * 14 = 88 answer : c
['a ) 76', 'b ) 55', 'c ) 88', 'd ) 21', 'e ) 11']
c
circumface(sqrt(divide(616, const_pi)))
divide(n0,const_pi)|sqrt(#0)|circumface(#1)
geometry
the ratio of boarders to day scholars at a school is 7 to 16 . however , after a few new students join the initial 560 boarders , the ratio changed to 1 to 2 , respectively . if no boarders became day scholars and vice versa , and no students left the school , how many boarders joined the school ?
let day scholars be ds and boarders be b given b / ds = 7 / 16 - - - > eq 1 initially b = 560 and x be new boarders joined then from eq 1 , we get ds = 16 / 7 * 560 = 1280 now , new ration is 560 + x / ds = 1 / 2 560 + x / 1280 = 1 / 2 = > 1120 + 2 x = 1280 = > 2 x = 160 = > x = 80 ans option d
a ) 48 , b ) 64 , c ) 70 , d ) 80 , e ) 84
d
subtract(divide(multiply(divide(560, 7), 16), divide(2, add(1, 2))), add(560, multiply(divide(560, 7), 16)))
add(n3,n4)|divide(n2,n0)|divide(n4,#0)|multiply(n1,#1)|add(n2,#3)|divide(#3,#2)|subtract(#5,#4)
other
when positive integer n is divided by positive integer p , the quotient is 18 , with a remainder of 6 . when n is divided by ( p + 2 ) , the quotient is 15 and the remainder is 3 . what is the value of n ?
n / p = 18 6 / p = 18 p + 6 n / ( p + 2 ) = 15 2 / ( p + 2 ) = 15 p + 30 + 3 solving these two equations we get p = 9 n = 168 answer is c .
a ) 151 , b ) 331 , c ) 168 , d ) 691 , e ) 871
c
multiply(18, divide(subtract(add(multiply(15, 2), 3), 6), subtract(18, 15)))
multiply(n2,n3)|subtract(n0,n3)|add(n4,#0)|subtract(#2,n1)|divide(#3,#1)|multiply(n0,#4)
general
in an examination , 25 % of total students failed in hindi , 35 % failed in english and 40 % in both . the percentage of these who passed in both the subjects is :
"pass percentage = 100 - ( 25 + 35 - 40 ) = 100 - 20 = 80 answer : d"
a ) 10 % , b ) 20 % , c ) 60 % , d ) 80 % , e ) 50 %
d
subtract(const_100, subtract(add(25, 35), 40))
add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|
general
population of a city in 20004 was 1400000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city at the end of the year 2007
"required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 2818075 / 2 e"
a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 2818075 / 2
e
multiply(1400000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100))))
divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)|
gain
the slant height of a cone is 15 cm and radius of the base is 3 cm , find the curved surface of the cone ?
"π * 15 * 3 = 141 answer : c"
a ) 26 , b ) 134 , c ) 141 , d ) 190 , e ) 28
c
multiply(multiply(const_pi, 3), 15)
multiply(n1,const_pi)|multiply(n0,#0)|
geometry
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 340 m long ?
"speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 340 = 700 m required time = 700 * 2 / 25 = 56 sec answer : a"
a ) 56 sec , b ) 42 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec
a
divide(360, multiply(subtract(45, 340), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
if ( 55 ^ 55 + 55 ) is divided by 56 , then the remainder is : ?
solution : ( x ^ n + 1 ) is divisible by ( x + 1 ) , when n is odd . . ' . ( 55 ^ 55 + 1 ) is divisible by ( 55 + 1 ) = 56 . when ( 55 ^ 55 + 1 ) + 54 is divided by 56 , the remainder is 54 . answer a
a ) 54 , b ) 55 , c ) 53 , d ) 56 , e ) 57
a
add(55, subtract(55, 56))
subtract(n0,n3)|add(n0,#0)
general
g ( x ) is defined as the product of all even integers k such that 0 < k ≤ x . for example , g ( 14 ) = 2 × 4 × 6 × 8 × 10 × 12 × 14 . if g ( a ) is divisible by 4 ^ 11 , what is the smallest possible value for a ?
"g ( a ) = 4 ^ 11 = 2 ^ 22 . so we have to find a product with atleast 22 2 ' s in it . in option 1 22 the total no of 2 ' s = [ 22 / 2 ] + [ 22 / 4 ] + [ 22 / 8 ] + [ 22 / 16 ] = 11 + 5 + 2 + 1 = 19 in option 2 24 the total no of 2 ' s = [ 24 / 2 ] + [ 24 / 4 ] + [ 24 / 8 ] + [ 24 / 16 ] = 12 + 6 + 3 + 1 = 22 . hence b"
a ) 22 , b ) 24 , c ) 28 , d ) 32 , e ) 44
b
multiply(2, 12)
multiply(n2,n7)|
general
the least common multiple of positive integer m and 3 - digit integer n is 690 . if n is not divisible by 3 and m is not divisible by 2 , what is the value of n ?
the lcm of n and m is 690 = 2 * 3 * 5 * 23 . m is not divisible by 2 , thus 2 goes to n n is not divisible by 3 , thus 3 goes to m . from above : n must be divisible by 2 and not divisible by 3 : n = 2 * . . . in order n to be a 3 - digit number it must take all other primes too : n = 2 * 5 * 23 = 230 . answer : b .
a ) 115 , b ) 230 , c ) 460 , d ) 575 , e ) 690
b
divide(690, 3)
divide(n1,n0)
general
if 5 % more is gained by selling an article for rs . 350 than by selling it for rs . 340 , the cost of the article is :
"let c . p . be rs . x . then , 5 % of x = 350 - 340 = 10 x / 20 = 10 = > x = 200 answer : c"
a ) rs . 50 , b ) rs . 160 , c ) rs . 200 , d ) rs . 225 , e ) rs . 325
c
divide(subtract(350, 340), divide(5, const_100))
divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)|
gain
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent f of the distribution is less than m + d ?
"d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , f = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d . d"
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
d
subtract(const_100, divide(subtract(const_100, 68), const_2))
subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)|
general
the difference between the local value and the face value of 9 in the numeral 62975148 is
"explanation : ( local value of 9 ) - ( face value of 9 ) = ( 900000 - 9 ) = 899991 b )"
a ) 899999 , b ) 899991 , c ) 899891 , d ) 899981 , e ) 899000
b
subtract(multiply(9, const_1000), 9)
multiply(n0,const_1000)|subtract(#0,n0)|
general
a man complete a journey in 10 hours . he travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km .
"( 1 / 2 ) x / 21 + ( 1 / 2 ) x / 24 = 10 x / 21 + x / 24 = 20 15 x = 168 x 20 x = [ 168 x 20 / 15 ] = 224 km answer is b"
a ) 220 km , b ) 224 km , c ) 230 km , d ) 234 km , e ) 255 km
b
multiply(const_2, divide(multiply(multiply(21, 24), 10), add(21, 24)))
add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)|
physics
the compound ratio of 3 : 4 , 3 : 2 and 4 : 5 ?
"3 / 4 * 3 / 2 * 4 / 5 = 9 / 10 = 9 : 10 answer : d"
a ) 1 : 9 , b ) 1 : 7 , c ) 1 : 2 , d ) 9 : 10 , e ) 1 : 4
d
divide(divide(multiply(3, 3), multiply(4, 2)), divide(multiply(3, 4), multiply(2, 5)))
multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)|
other
the true discount on a bill of rs . 2460 is rs . 360 . what is the banker ' s discount ?
"explanation : f = rs . 2460 td = rs . 360 pw = f - td = 2460 - 360 = rs . 2100 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2100 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 . 2460 for unexpired time = 360 / 2100 × 2460 = 0.17 × 2460 = rs . 422 answer : option b"
a ) rs . 432 , b ) rs . 422 , c ) rs . 412 , d ) rs . 442 , e ) none of these
b
multiply(divide(360, subtract(2460, 360)), 2460)
subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)|
gain
a number is doubled and 5 is added . if the resultant is trebled , it becomes 111 . what is that number ?
"explanation : let the number be x . therefore , 3 ( 2 x + 5 ) = 111 6 x + 15 = 111 6 x = 96 x = 16 answer : d"
a ) 12 , b ) 29 , c ) 27 , d ) 16 , e ) 99
d
divide(subtract(111, multiply(const_3, 5)), multiply(const_3, const_2))
multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)|
general
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 700 more than d , what is b ' s share ?
"let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 700 x = 700 . b ' s share = rs . 2 x = rs . ( 2 x 700 ) = rs . 1400 . answer = b"
a ) rs . 500 , b ) rs . 1400 , c ) rs . 2000 , d ) rs . 2500 , e ) none of the above
b
multiply(multiply(subtract(4, 3), 700), 3)
subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)|
general
if the simple interest on a sum of money for 5 years at 9 % per annum is rs . 800 , what is the compound interest on the same sum at the rate and for the same time ?
"sum = ( 800 * 100 ) / ( 5 * 9 ) = rs . 1 , 777.78 c . i . on rs . rs . 1 , 777.78 for 5 years at 9 % = rs . 2 , 735.33 . = rs . 2 , 735.33 - 1 , 777.78 = rs . 957.55 answer : b"
a ) rs . 947.55 , b ) rs . 957.55 , c ) rs . 857.55 , d ) rs . 657.55 , e ) rs . 357.55
b
subtract(add(add(divide(multiply(divide(800, multiply(divide(9, const_100), 5)), 9), const_100), divide(800, multiply(divide(9, const_100), 5))), divide(multiply(add(divide(multiply(divide(800, multiply(divide(9, const_100), 5)), 9), const_100), divide(800, multiply(divide(9, const_100), 5))), 9), const_100)), divide(800, multiply(divide(9, const_100), 5)))
divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|
gain
find the difference between c . i and s . i on a sum of money rs . 2000 for 2 years . at 4 % p . a
p 9 r / 100 ) ^ 2 = 2000 ( 4 / 100 ) ^ 2 = rs . 3.20 answer : b
a ) rs . 2.20 , b ) rs . 3.20 , c ) rs . 4.20 , d ) rs . 5.20 , e ) rs . 3.25
b
subtract(subtract(multiply(multiply(2000, add(const_1, divide(4, const_100))), add(const_1, divide(4, const_100))), 2000), multiply(2, multiply(2000, divide(4, const_100))))
divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n0,#1)|multiply(n1,#2)|multiply(#1,#3)|subtract(#5,n0)|subtract(#6,#4)
general
if 5 machines can produce 20 units in 10 hours , how long would it take 10 machines to produce 100 units ?
"5 machines would produce 100 units in 50 hours . increasing the amount of machines by 2 would mean dividing 50 hours by 2 . 50 / 2 = 25 answer : a"
a ) 25 , b ) 30 , c ) 35 , d ) 24 , e ) 96
a
divide(100, multiply(divide(divide(10, 10), 5), 20))
divide(n3,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)|
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 a ' 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 , the ratio of the work done by a : b : c when they work together will be equal to 1 / 6 : 1 / 8 : 1 / 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 / 6 : 24 / 8 : 24 / 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 2340 will be 4 : 3 : 2 . hence , a ' s share will be 4 * 2340 / 9 = 1040 correct choice is ( c )"
a ) $ 1100 , b ) $ 520 , c ) $ 1040 , 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
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.00 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ?
"a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 – 6 = 666 * 1.0 = 666 answer : a"
a ) s . 666 , b ) s . 1140 , c ) s . 999 , d ) s . 1085 , e ) s . 1020
a
multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.00), 3)
multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)|
physics
the pinedale bus line travels at an average speed of 60 km / h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 5 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ?
"number of stops in an hour : 60 / 5 = 12 distance between stops : 60 / 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 5 = 25 km imo , correct answer is ` ` a . ' '"
a ) 25 km , b ) 30 km , c ) 40 km , d ) 50 km , e ) 60 km
a
multiply(60, divide(multiply(5, 5), 60))
multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|
physics
a positive number x is multiplied by 10 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ?
"sq rt ( 10 x / 3 ) = x = > 10 x / 3 = x ^ 2 = > x = 10 / 3 ans - e"
a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 2 / 3 , e ) 10 / 3
e
divide(10, 3)
divide(n0,n1)|
general
if 90 % of a = 30 % of b and b = c % of a , then the value of c is ?
"answer ∵ 90 a / 100 = 30 b / 100 = ( 30 / 100 ) x ac / 100 ∴ c = 100 x ( 100 / 30 ) x ( 90 / 100 ) = 300 correct option : d"
a ) 900 , b ) 800 , c ) 600 , d ) 300 , e ) none
d
divide(30, 90)
divide(n1,n0)|
gain
in a group of dogs and people , the number of legs was 28 more than twice the number of heads . how many dogs were there ? [ assume none of the people or dogs is missing a leg . ]
if there were only people , there would be exactly twice the number of legs as heads . each dog contributes two extra legs over and above this number . so 28 extra legs means 14 dogs . correct answer d
a ) 4 , b ) 7 , c ) 12 , d ) 14 , e ) 28
d
divide(28, const_2)
divide(n0,const_2)
general
in a company , 50 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ?
"the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 50 % of all employees are women , so non - union women are 50 % - 18 % = 32 % 40 % of all employees are non - union . the percent of non - union employees who are women is 32 % / 40 % = 80 % the answer is b ."
a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 %
b
multiply(const_100, divide(subtract(subtract(const_100, 50), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60)))
divide(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(n1,#0)|subtract(n1,#3)|subtract(#1,#4)|divide(#5,#2)|multiply(#6,const_100)|
gain
9 log 9 ( 4 ) = ?
"exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 4 ) = 4 correct answer d"
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
d
divide(log(multiply(9, 9)), log(const_10))
log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)|
other
the average of runs of a cricket player of 20 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 5 ?
"average = total runs / no . of innings = 32 so , total = average x no . of innings = 32 * 20 = 640 now increase in avg = 4 runs . so , new avg = 32 + 5 = 37 runs total runs = new avg x new no . of innings = 37 * 21 = 777 runs made in the 11 th inning = 777 - 640 = 137 answer : d"
a ) 96 , b ) 106 , c ) 128 , d ) 137 , e ) 122
d
subtract(multiply(add(20, const_1), add(5, 32)), multiply(20, 32))
add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
general
yearly subscription to professional magazines cost a company $ 840.00 . to make a 30 % cut in the magazine budget , how much less must be spent ?
"total cost 840 840 * 30 / 100 = 252 so the cut in amount is 252 the less amount to be spend is 840 - 252 = 588 answer : d"
a ) 654 , b ) 655 , c ) 656 , d ) 588 , e ) 658
d
multiply(divide(subtract(const_100, 30), const_100), 840.00)
subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1)|
general
the area of a square is 4900 sq cm . find the ratio of the breadth and the length of a rectangle whose length is same the side of the square and breadth is 20 cm less than the side of the square ?
"let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4900 a = ( 4900 ) ^ 1 / 2 = 70 l = a and b = a - 20 b : l = a - 20 : a = 50 : 70 = 5 : 7 answer : d"
a ) 5 : 8 , b ) 5 : 10 , c ) 5 : 13 , d ) 5 : 7 , e ) 5 : 16
d
divide(subtract(sqrt(4900), 20), multiply(sqrt(4900), const_2))
sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)|
geometry
a man walking at a rate of 10 km / hr crosses a bridge in 18 minutes . the length of the bridge is ?
"speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 10 minutes = 50 / 18 * 18 * 60 = 3000 m answer is b"
a ) 1521 , b ) 3000 , c ) 1667 , d ) 1254 , e ) 1112
b
multiply(divide(multiply(10, const_1000), const_60), 18)
multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|
gain
the inner circumference of a circular race track , 14 m wide , is 440 m . find radius of the outer circle .
let inner radius be r metres . then , 2 ( 22 / 7 ) r = 440 = r = ( 440 x ( 7 / 44 ) ) = 70 m . radius of outer circle = ( 70 + 14 ) m = 84 m . answer is a
['a ) 84', 'b ) 86', 'c ) 82', 'd ) 80', 'e ) none of them']
a
add(14, divide(divide(440, const_pi), const_2))
divide(n1,const_pi)|divide(#0,const_2)|add(n0,#1)
physics
r is the set of positive even integers less than 20 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ?
squares < 20 { 1,4 , 9,16 } s = { 1,4 , 9,16 } r = { 2 , . . . . . 18 } hence c .
['a ) none', 'b ) two', 'c ) three', 'd ) five', 'e ) seven']
c
subtract(const_4, const_1)
subtract(const_4,const_1)
physics
elena ’ s bread recipe calls for 3 ounces of butter for each 4 cups of flour used . she needs to make 7 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ?
"solving through algebra route : 3 b + 4 f = x amount if we multiply this equation with 7 we get : 21 b + 28 f = 7 x therefore , we got 21 ounces of butter and 7 x amount of quantity when we use 28 ounces of floor . ans : e"
a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 28
e
multiply(4, 7)
multiply(n1,n2)|
general
find the area , diameter = 13 m .
"diameter = 13 meter . radius = diameter / 2 . = 13 / 2 . = 6.5 meter . area of a circle = ï € r 2 . here , pi ( ï € ) = 3.14 meter , radius ( r ) = 6.5 area of a circle = 3.14 ã — 6.5 ã — 6.5 . = 3.14 ã — 42.25 . = 132.78 square meter answer : c"
a ) 132.00 square meter , b ) 132.04 square meter , c ) 132.78 square meter , d ) 132.24 square meter , e ) 113.43 square meter
c
circle_area(divide(13, const_2))
divide(n0,const_2)|circle_area(#0)|
physics
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 45 , find the other ?
"other number = 12 * 600 / 45 = 160 answer is b"
a ) 100 , b ) 160 , c ) 120 , d ) 200 , e ) 150
b
multiply(12, 45)
multiply(n0,n2)|
physics
in a can , there is a mixture of milk and water in the ratio 4 : 3 . if the can is filled with an additional 8 liters of milk , the can would be full and the ratio of milk and water would become 2 : 1 . find the capacity of the can ?
"let c be the capacity of the can . ( 4 / 7 ) * ( c - 8 ) + 8 = ( 2 / 3 ) * c 12 c - 96 + 168 = 14 c 2 c = 72 c = 36 the answer is b ."
a ) 40 , b ) 36 , c ) 32 , d ) 28 , e ) 24
b
add(add(multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 4))), 4), 8), multiply(divide(multiply(1, 8), subtract(multiply(3, 2), multiply(1, 4))), 3))
multiply(n2,n4)|multiply(n1,n3)|multiply(n0,n4)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)|multiply(n1,#4)|add(n2,#5)|add(#7,#6)|
general
according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 272 6 ounces servings of orange juice ?
"its b . total juice rquired = 272 * 6 = 1632 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1632 / 48 = 34 . answer b"
a ) a ) 25 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100
b
divide(divide(multiply(272, 6), 12), const_4)
multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)|
general
to produce an annual income of rs . 1200 from a 12 % stock at 90 , the amount of stock needed is :
solution for an income of rs . 12 , stock needed = rs . 100 . for an income of rs . 1200 , stock needed = rs . 100 / 12 x 1200 = 10000 answer a
a ) rs . 10,000 , b ) rs . 10,800 , c ) rs . 14,400 , d ) rs . 16,000 , e ) none of these
a
subtract(const_100, 90)
subtract(const_100,n2)
gain
if the least common addition of two prime numbers x and y is 10 , where x > y , then the value of 2 x + y is
"( x + y ) = 10 and both x an y are prime . the only values of x and y can be 7 and 3 ( x = 7 and y = 3 ) 2 x + y = 2 * 7 + 3 = 17 correct option : d"
a ) 7 , b ) 9 , c ) 14 , d ) 17 , e ) 21
d
add(add(multiply(10, const_2), divide(10, 2)), add(const_3, const_3))
add(const_3,const_3)|divide(n0,n1)|multiply(n0,const_2)|add(#1,#2)|add(#3,#0)|
general
what is the length of a bridge ( in meters ) , which a train 156 meters long and travelling at 45 km / h can cross in 40 seconds ?
"speed = 45 km / h = 45000 m / 3600 s = 25 / 2 m / s in 40 seconds , the train can travel 25 / 2 * 40 = 500 meters 500 = length of train + length of bridge length of bridge = 500 - 156 = 344 meters the answer is d ."
a ) 308 , b ) 320 , c ) 332 , d ) 344 , e ) 356
d
subtract(multiply(multiply(45, const_0_2778), 40), 156)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics
a store has 10 bottles of juice , including 6 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 3 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought .
"the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 3 bottles of apple juice is 6 c 3 * 4 c 3 = 20 * 4 = 80 p ( selling 3 bottles of apple juice ) = 80 / 210 = 8 / 21 the answer is d ."
a ) 2 / 11 , b ) 4 / 15 , c ) 6 / 17 , d ) 8 / 21 , e ) 10 / 25
d
divide(choose(6, 3), choose(10, 6))
choose(n1,n3)|choose(n0,n1)|divide(#0,#1)|
probability
if 10 men can reap 120 acres of land in 15 days , how many acres of land can 20 men reap in 30 days ?
"10 men 120 acres 15 days 20 men ? 30 days 120 * 20 / 10 * 30 / 15 120 * 2 * 2 120 * 4 = 480 answer : e"
a ) 120 , b ) 240 , c ) 360 , d ) 1200 , e ) 480
e
multiply(120, multiply(divide(20, 10), divide(30, 15)))
divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2)|
physics
the profit earned by selling an article for 852 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ?
"let the profit or loss be x and 852 – x = 448 + x or , x = 404 ⁄ 2 = 202 \ cost price of the article = 852 – x = 448 + x = 650 \ sp of the article = 650 × 150 ⁄ 100 = 975 answer b"
a ) 960 , b ) 975 , c ) 1,200 , d ) 920 , e ) none of these
b
multiply(divide(add(852, 448), const_2), add(const_1, divide(50, const_100)))
add(n0,n1)|divide(n2,const_100)|add(#1,const_1)|divide(#0,const_2)|multiply(#2,#3)|
gain
cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 6 % sugar , what should be the ratio of cereal a to cereal b , by weight ?
"( 10 / 100 ) a + ( 2 / 100 ) b = ( 6 / 100 ) ( a + b ) 4 a = 4 b = > a / b = 1 / 1 answer is d ."
a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 1 , e ) 1 : 3
d
divide(subtract(6, 2), subtract(10, 6))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|
general
zachary is helping his younger brother , sterling , learn his multiplication tables . for every question that sterling answers correctly , zachary gives him 3 pieces of candy . for every question that sterling answers incorrectly , zachary takes away two pieces of candy . after 7 questions , if sterling had answered 2 more questions correctly , he would have earned 31 pieces of candy . how many of the 7 questions did zachary answer correctly ?
"i got two equations : 3 x - 2 y = 25 x + y = 7 3 x - 2 ( 7 - x ) = 25 3 x - 14 + 2 x = 25 5 x = 39 x = 7.8 or between 7 and 8 . ( ans b )"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
b
divide(add(subtract(multiply(2, subtract(7, 2)), multiply(3, 2)), 31), add(3, 2))
add(n0,n2)|multiply(n0,n2)|subtract(n1,n2)|multiply(n2,#2)|subtract(#3,#1)|add(n3,#4)|divide(#5,#0)|
general
find the least number which when divided by 35 and 11 leaves a remainder of 1 in each case .
"explanation : the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 35 , 11 ) + 1 = 386 . answer : e"
a ) 366 , b ) 236 , c ) 367 , d ) 365 , e ) 386
e
add(1, lcm(35, 11))
lcm(n0,n1)|add(n2,#0)|
general
the speed of a car is 90 km in the first hour and 75 km in the second hour . what is the average speed of the car ?
"s = ( 90 + 75 ) / 2 = 82.5 kmph b"
a ) 89 kmph , b ) 82.5 kmph , c ) 75 kmph , d ) 65 kmph , e ) 77 kmph
b
divide(add(90, 75), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
a larger cube has 64 cubic inch as a volume and in the cube there are 64 smaller cubes such that their volume is 1 cubic inch . what is the difference between the surface areas ’ sum of the 64 smaller cubes and the surface area of the larger cube , in square inch ?
"volume of larger cube = 64 = 4 ^ 3 side of larger cube = 4 volume of smaller cube = 1 - - > side of smaller cube = 1 surface area of larger cube = 6 * 4 ^ 2 = 96 surface area of 27 smaller cubes = 64 * 6 * 1 = 384 difference = 384 - 96 = 288 answer : d"
a ) 54 , b ) 64 , c ) 81 , d ) 288 , e ) 120
d
subtract(multiply(surface_cube(cube_edge_by_volume(1)), 64), surface_cube(cube_edge_by_volume(64)))
cube_edge_by_volume(n2)|cube_edge_by_volume(n0)|surface_cube(#0)|surface_cube(#1)|multiply(n0,#2)|subtract(#4,#3)|
geometry
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 8 hours to fill the tank . the leak can drain all the water of the tank in ?
"work done by the tank in 1 hour = ( 1 / 2 - 2 1 / 8 ) = 1 / 34 leak will empty the tank in 34 hrs . answer : d"
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 34 hr , e ) 36 hr
d
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 8), 1), 8))))
divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)|
physics
two buses each 3125 m long are running in opposite directions on parallel roads . their speeds are 40 km / hr and 35 km / hr respectively . find the time taken by the slower bus to pass the driver of the faster one ?
relative speed = 40 + 35 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 3125 + 3125 = 6250 m . required time = 6250 * 6 / 125 = 50 sec . answer : a
a ) 50 sec , b ) 66 sec , c ) 48 sec , d ) 55 sec , e ) 45 sec
a
divide(divide(3125, divide(multiply(add(40, 35), const_1000), const_3600)), const_3)
add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|divide(#3,const_3)
physics
if the perimeter of a rectangular garden is 500 m , its length when its breadth is 100 m is ?
"2 ( l + 100 ) = 500 = > l = 150 m answer : e"
a ) 299 m , b ) 777 m , c ) 200 m , d ) 167 m , e ) 150 m
e
subtract(divide(500, const_2), 100)
divide(n0,const_2)|subtract(#0,n1)|
physics
a man goes from a to b at a speed of 15 kmph and comes back to a at a speed of 12 kmph . find his average speed for the entire journey ?
distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 15 ) + ( d / 12 ] = ( 2 d ) / [ 9 d / 60 ) = > 13.3 kmph . answer : b
a ) 13.7 kmph , b ) 13.3 kmph , c ) 13.8 kmph , d ) 13.9 kmph , e ) 12.3 kmph
b
divide(add(15, 12), const_2)
add(n0,n1)|divide(#0,const_2)
physics
a cube is divided into 729 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that the cube is coloured with green colour on one set of adjacent faces , red on the other set of adjacent faces , blue on the third set . so , how many cubelets are there which are painted with exactly one colour ?
"total cubes created are 729 so a plane of big cube has 9 * 9 cubes out of that ( n - 2 ) * ( n - 2 ) = 7 * 7 = 49 are painted only one side and a cube has six sides = 6 * 49 = 294 answer : b"
a ) 293 , b ) 294 , c ) 295 , d ) 296 , e ) 298
b
divide(subtract(729, multiply(multiply(const_4, const_2), const_3)), const_2)
multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)|
geometry
in a mixed college 160 students are there in one class . out of this 160 students 3 / 4 students are girls . how many boys are there ?
"total number of students : 160 total girls : 160 * 3 / 4 = 120 total boys : 160 - 120 = 40 answer is a"
a ) 40 , b ) 60 , c ) 80 , d ) 120 , e ) 140
a
multiply(divide(160, 4), const_3)
divide(n0,n3)|multiply(#0,const_3)|
general
the perimeter of a triangle is 60 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle
"explanation : area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 60 / 2 = 35 cm 2 answer : option d"
a ) a ) 72 , b ) b ) 828 , c ) c ) 729 , d ) d ) 75 , e ) e ) 35
d
triangle_area(2.5, 60)
triangle_area(n0,n1)|
geometry
the product z of two prime numbers is between 10 and 30 . if one of the prime numbers is greater than 2 but less than 6 and the other prime number is greater than 6 but less than 24 , then what is z ?
"the smallest possible product is 21 which is 3 * 7 . all other products are too big . the answer is a ."
a ) 21 , b ) 15 , c ) 14 , d ) 10 , e ) 6
a
multiply(subtract(10, const_4), const_3)
subtract(n0,const_4)|multiply(#0,const_3)|
general
how many integers from 0 to 44 , inclusive , have a remainder of 1 when divided by 3 ?
"explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 43 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term of an ap , a is the first term and d is the common difference . so , 43 = 1 + ( n - 1 ) 3 or , ( n - 1 ) 3 = 42 or , n - 1 = 12 or , n = 13 a"
a ) 13 , b ) 16 , c ) 17 , d ) 18 , e ) 19
a
divide(44, const_10)
divide(n1,const_10)|
general
he total marks obtained by a student in physics , chemistry and mathematics is 150 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ?
"let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 150 + p c + m = 150 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 150 / 2 = 75 . answer : c"
a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 95
c
divide(150, const_2)
divide(n0,const_2)|
general