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a ) 1 / 4 , b ) 3 / 8 , c ) 9 / 16 , d ) 1 / 8 , e ) 16 / 9
d
multiply(divide(3, 7), multiply(divide(7, 8), divide(1, 3)))
what is 3 / 7 of 7 / 8 of 1 / 3 ?
3 / 7 * 7 / 8 * 1 / 3 = 1 / 8 answer : d
a = 3 / 7 b = 7 / 8 c = 1 / 3 d = b * c e = a * d
a ) 3 , b ) 13 , c ) 20 , d ) 33 , e ) 43
c
subtract(multiply(multiply(multiply(const_2, const_2), 5), 3), divide(2400, multiply(multiply(multiply(const_2, const_2), 5), 3)))
the least number which should be added to 2400 so that the sum is exactly divisible by 5 , 6,4 and 3 is :
solution l . c . m . of 5 , 6,4 and 3 = 60 . on dividing 2400 by 60 , the remainder is 40 . ∴ number to be added = ( 60 - 40 ) = 20 . answer c
a = 2 * 2 b = a * 5 c = b * 3 d = 2 * 2 e = d * 5 f = e * 3 g = 2400 / f h = c - g
a ) 8400 , b ) 8200 , c ) 6000 , d ) 5000 , e ) 4000
b
divide(82, divide(subtract(7, 6), const_100))
in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 82 more candidates got selected than a . what was the number of candidates appeared from each state ?
"state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 82 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 82 = > total appeared candidates in state b = 82 x 100 = 8200 = > total appeared candidates in state a = total appeared candidates in state b = 8200"
a = 7 - 6 b = a / 100 c = 82 / b
a ) 28 sec , b ) 16 sec , c ) 34 sec , d ) 18 sec , e ) 17 sec
c
divide(add(240, 100), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))
a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 100 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 100 = 340 m . time taken = 340 / 10 = 34 sec . answer : c"
a = 240 + 100 b = 45 - 9 c = 10 / 2 d = 45 - 9 e = d / 2 f = c / e g = b * f h = a / g
a ) 1888 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2991
c
divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 8720), divide(multiply(15, 8720), const_100)), const_2)
in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 8720 votes were polled , then how many valid votes did b get ?
"let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 8720 32.5 k = 32.5 k / 100 k * 8720 = 2834 the number of valid votes polled in favour of b is 2834 . answer : c"
a = 20 / 100 b = 1 - a c = b * 8720 d = 15 * 8720 e = d / 100 f = c - e g = f / 2
a ) $ 1200 , b ) $ 1005.8 , c ) $ 555.6 , d ) $ 1009.2 , e ) $ 1490.7
c
multiply(2500, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8)))))
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 $ 2500 , what will be c ' s share of the earnings ?
"the dollars earned will be in the same ratio as amount of work done 1 day work of c is 1 / 12 ( or 2 / 24 ) 1 day work of the combined workforce is ( 1 / 6 + 1 / 8 + 1 / 12 ) = 9 / 24 c ' s contribution is 2 / 9 of the combined effort translating effort to $ = 2 / 9 * 2500 = $ 555.6"
a = 1/(8) b = 1/(12) c = 1/(6) d = 1/(8) e = c + d f = b + e g = a / f h = 2500 * g
a ) 53 , b ) 48 , c ) 59 , d ) 47 , e ) 50
d
subtract(negate(37), multiply(subtract(29, 31), divide(subtract(29, 31), subtract(23, 29))))
23 , 29 , 31 , 37 , 41 , 43 , ( . . . )
"all are prime numbers in their order , starting from 23 hence , next number is 47 answer is d"
a = negate - (
a ) 299 , b ) 666 , c ) 778 , d ) 900 , e ) 977
d
multiply(add(5, 6), const_100)
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 81 . how much was lent at 5 % ?
"( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 81 5 x / 100 + 90 – 6 x / 100 = 81 x / 100 = 9 = > x = 900 answer : d"
a = 5 + 6 b = a * 100
a ) 164.19 , b ) 164.12 , c ) 170.6 , d ) 167.1 , e ) 165.11
c
subtract(subtract(multiply(1000, power(add(divide(16, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(16, const_100)), 4))
what will be the difference between simple and compound interest at 16 % per annum on a sum of rs . 1000 after 4 years ?
"s . i . = ( 1000 * 16 * 4 ) / 100 = rs . 640 c . i . = [ 1000 * ( 1 + 16 / 100 ) 4 - 1000 ] = rs . 810.6 difference = ( 810.6 - 640 ) = rs . 170.6 answer : c"
a = 16 / 100 b = a + 1 c = b ** 4 d = 1000 * c e = d - 1000 f = 16 / 100 g = 1000 * f h = g * 4 i = e - h
a ) 2.5 sec , b ) 2.95 sec , c ) 1.3 sec , d ) 8.75 sec , e ) 1.85 sec
d
divide(350, multiply(144, const_0_2778))
in what time will a train 350 m long cross an electric pole , it its speed be 144 km / hr ?
"speed = 144 * 5 / 18 = 40 m / sec time taken = 100 / 40 = 8.75 sec . answer : d"
a = 144 * const_0_2778 b = 350 / a
a ) 440 , b ) 540 , c ) 640 , d ) 722 , e ) 250
d
divide(multiply(530, 5), divide(add(multiply(3, 3), 2), 3))
an aeroplane covers a certain distance of 530 kmph in 5 hours . to cover the same distance in 3 2 / 3 hours , it must travel at a speed of
"speed of aeroplane = 530 kmph distance travelled in 5 hours = 530 * 5 = 2650 km speed of aeroplane to acver 2650 km in 11 / 3 = 2650 * 3 / 11 = 722 km answer d"
a = 530 * 5 b = 3 * 3 c = b + 2 d = c / 3 e = a / d
a ) 2 / 7 , b ) 3 / 7 , c ) 3 / 14 , d ) 5 / 28 , e ) 9 / 28
c
multiply(divide(4, add(4, 4)), divide(subtract(4, const_1), subtract(add(4, 4), const_1)))
there are 4 red books and 4 blue books on a shelf . if 2 books are selected at random from the shelf , what is the probability that both books selected are red books ?
the number of ways of choosing 2 books from the shelf is 8 c 2 = 28 . the number of ways of choosing 2 red books is 4 c 2 = 6 . p ( 2 red books ) = 6 / 28 = 3 / 14 . the answer is c .
a = 4 + 4 b = 4 / a c = 4 - 1 d = 4 + 4 e = d - 1 f = c / e g = b * f
a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 11 / 16 , e ) 15 / 16
e
subtract(const_1, power(divide(const_1, 2), 4))
on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow ’ s neck . what is the probability that the rancher will be able to place a lasso around a cow ’ s neck at least once in 4 attempts ?
p ( missing all 4 ) = ( 1 / 2 ) ^ 4 = 1 / 16 p ( success on at least one attempt ) = 1 - 1 / 16 = 15 / 16 the answer is e .
a = 1 / 2 b = a ** 4 c = 1 - b
a ) 2 , b ) 7 , c ) 9 , d ) 10 , e ) 15
d
divide(add(222, 43), 17)
a no . when divided by 222 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ?
"222 + 43 = 265 / 17 = 10 ( remainder ) d"
a = 222 + 43 b = a / 17
a ) 1 : 8 , b ) 1 : 7 , c ) 1 : 6 , d ) 1 : 5 , e ) 1 : 4
b
multiply(subtract(divide(4, 7), divide(const_1, const_2)), const_2)
the ratio of the arithmetic mean of two numbers to one of the numbers is 4 : 7 . what is the ratio of the smaller number to the larger number ?
"for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 4 : 7 , thus the smaller number must have a ratio of 1 . the ratio of the smaller number to the larger number is 1 : 7 . the answer is b ."
a = 4 / 7 b = 1 / 2 c = a - b d = c * 2
a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0005 , e ) 0.0002
a
divide(5, 1,000)
when magnified 1,000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 5 centimeter . the actual diameter of the tissue , in centimeters , is
"it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 5 then x = 5 / 1000 = 0.005 . the answer is a"
a = 5 / 1
a ) 50 , b ) 99 , c ) 88 , d ) 77 , e ) 21
a
divide(360, multiply(subtract(45, 265), const_0_2778))
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 265 m long ?
": speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 265 = 625 m required time = 625 * 2 / 25 = 50 sec answer : a"
a = 45 - 265 b = a * const_0_2778 c = 360 / b
a ) rs . 105 , b ) rs . 150 , c ) rs . 146 , d ) rs . 135 , e ) none of these
a
subtract(multiply(add(divide(add(4, divide(3, 4)), const_100), 1), divide(10500, add(divide(add(divide(1, 2), 1), const_100), 1))), 10500)
if the sales tax be reduced from 4 ( 1 / 2 ) % to 3 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 10500 ?
"explanation : required difference = ( 4 ( 1 / 2 ) of rs . 10500 ) - ( 3 ( 1 / 2 ) of rs . 10500 ) = ( 9 / 2 – 7 / 2 ) % of rs . 10500 = ( 1 ) x ( 1 / 100 ) x 10500 = rs . 105 answer a"
a = 3 / 4 b = 4 + a c = b / 100 d = c + 1 e = 1 / 2 f = e + 1 g = f / 100 h = g + 1 i = 10500 / h j = d * i k = j - 10500
a ) 2050 , b ) 2350 , c ) 2650 , d ) 2950 , e ) 3250
e
add(divide(subtract(subtract(151, 1), add(99, 1)), 2), 1)
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 151 ?
"100 + 102 + . . . + 150 = 100 * 26 + ( 2 + 4 + . . . + 50 ) = 100 * 26 + 2 * ( 1 + 2 + . . . + 25 ) = 100 * 26 + 2 ( 25 ) ( 26 ) / 2 = 100 * 26 + 25 * 26 = 125 ( 26 ) = 3250 the answer is e ."
a = 151 - 1 b = 99 + 1 c = a - b d = c / 2 e = d + 1
a ) 50 , b ) 70 , c ) 100 , d ) 130 , e ) 140
d
subtract(subtract(490, divide(subtract(multiply(4, divide(490, 7)), multiply(3, divide(multiply(4, divide(490, 7)), add(3, 1)))), divide(1, 3))), multiply(3, divide(multiply(4, divide(490, 7)), add(3, 1))))
in an apartment building that has 490 units , 4 out of every 7 units are currently rented , including 1 / 3 of the one - bedroom units . if , of the rented apartments , there is a 6 : 1 ratio of two - bedroom units to one - bedroom units , and the building only consists of two - bedroom and one - bedroom units , how many two - bedroom units are not rented ?
"rented = 4 / 7 * 490 = 280 . rented two - bedroom = 6 / 7 * 280 = 240 ; rented one - bedroom = 1 / 7 * 280 = 40 ( or 280 - 240 = 40 ) . rented one - bedroom units are 1 / 3 of all one - bedroom : 1 / 3 * { all one bedroom } = 40 - - > { all one - bedroom } = 120 { all two - bedroom } = 490 - 120 = 370 . two - bedroom not rented = 370 - 240 = 130 answer : d ."
a = 490 / 7 b = 4 * a c = 490 / 7 d = 4 * c e = 3 + 1 f = d / e g = 3 * f h = b - g i = 1 / 3 j = h / i k = 490 - j l = 490 / 7 m = 4 * l n = 3 + 1 o = m / n p = 3 * o q = k - p
a ) 45 , b ) 60 , c ) 72 , d ) 80 , e ) 100
c
divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2))
if six machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ?
"using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 6 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 4 = 10 / 3 * x x = 6 / 5 hour so 72 minutes answer : c"
a = 3 / 5 b = 3 / 4 c = b / 5 d = c / 30 e = d * 2 f = a / e
a ) 15 min , b ) 20 min , c ) 25 min , d ) 30 min , e ) 35 min
d
divide(30, 1)
a fill pipe can fill 1 / 2 of cistern in 30 minutes . in how many minutes , it can fill 1 / 2 of the cistern ?
"required time = 30 * 2 * 1 / 2 = 30 minutes answer is d"
a = 30 / 1
a ) 0.47 , b ) 47 , c ) 470 , d ) 4700 , e ) none
c
multiply(4.7, add(add(13.26, 9.43), 77.31))
the value of ( 4.7 × 13.26 + 4.7 × 9.43 + 4.7 × 77.31 ) is
"solution given expression = 4.7 × ( 13.26 + 9.43 + 77.31 ) = 4.7 × 100 = 470 . answer c"
a = 13 + 26 b = a + 77 c = 4 * 7
a ) 11 , b ) 13 , c ) 20 3 / 17 , d ) 14 , e ) none of these
b
inverse(add(divide(const_1, 23), divide(const_1, multiply(add(divide(30, const_100), const_1), 23))))
a is 30 % more efficient than b . how much time will they working together take to complete a job which a alone could have done in 23 days ?
explanation : ratio of times taken by a and b = 100 : 130 = 10 : 13 suppose b takes x days to work the 10 : 13 : : 23 : x = > x = 23 x 13 / 10 = 299 / 10 a ’ s one day work = 1 / 23 and b ’ s is 10 / 299 then a + b ‘ s work = 1 / 23 + 10 / 299 = 23 / 299 = 1 / 13 = > 13 days answer : option b
a = 1 / 23 b = 30 / 100 c = b + 1 d = c * 23 e = 1 / d f = a + e g = 1/(f)
a ) s . 800 , b ) s . 810 , c ) s . 815 , d ) s . 900 , e ) s . 920
c
multiply(2445, divide(1, 3))
a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2445 ?
"2445 * 1 / 3 = 815 option c"
a = 1 / 3 b = 2445 * a
a ) 50 km , b ) 76 km , c ) 18 km , d ) 60 km , e ) 97 km
d
multiply(10, divide(20, subtract(15, 10)))
if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is ?
"let the actual distance traveled be x km . then , x / 10 = ( x + 20 ) / 15 5 x - 300 = > x = 60 km . answer : d"
a = 15 - 10 b = 20 / a c = 10 * b
a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55
b
divide(add(40, 40), add(divide(40, 15), divide(40, 30)))
gwen drove an average speed of 15 miles per hour for the first 40 miles of a tripthen at a average speed of 30 miles / hr for the remaining 40 miles of the trip if she made no stops during the trip what was gwen ' s avg speed in miles / hr for the entire trip
avg . speed = total distance / total time total distance = 80 miles total time = 40 / 15 + 40 / 30 = 4 avg . speed = 20 . answer - b
a = 40 + 40 b = 40 / 15 c = 40 / 30 d = b + c e = a / d
a ) 1100 yards , b ) 1200 yards , c ) 1300 yards , d ) 1400 yards , e ) 1000 yards
b
multiply(300, const_4)
a boat m leaves shore a and at the same time boat b leaves shore b . they move across the river . they met at 500 yards away from a and after that they met 300 yards away from shore b without halting at shores . find the distance between the shore a & b .
"if x is the distance , a is speed of a and b is speed of b , then ; 500 / a = ( x - 500 ) / b and ( x + 300 ) / a = ( 2 x - 300 ) / b , solving it , we get x = 1200 yards answer : b"
a = 300 * 4
a ) 10 ^ 7 , b ) 10 ^ 8 , c ) 10 ^ 9 , d ) 10 ^ 10 , e ) 10 ^ 11
b
multiply(multiply(78, 48), add(add(multiply(multiply(const_0_25, const_1000), const_100), multiply(add(const_3, const_4), const_10)), const_3))
78 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,075 bacteria each . how many bacteria are there approximately ?
"78 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,075 bacteria each . how many bacteria are there approximately ? a . 10 ^ 7 b . 10 ^ 8 c . 10 ^ 9 d . 10 ^ 10 e . 10 ^ 11 - > due to approximately , 78 = 80 , 48 = 50 , 25,075 = 25,000 are derived , which makes ( 78 ) ( 48 ) ( 25,075 ) = ( 80 ) ( 50 ) ( 25,000 ) = 10 ^ 8 . the answer is b ."
a = 78 * 48 b = const_0_25 * 1000 c = b * 100 d = 3 + 4 e = d * 10 f = c + e g = f + 3 h = a * g
a ) 120 / 9 , b ) 140 / 9 , c ) 150 / 9 , d ) 190 / 9 , e ) 170 / 9
d
divide(subtract(multiply(10, 1200), multiply(820, 10)), subtract(1000, 820))
the average salary / head of all the workers in a workshop is rs . 1000 , if the average salary / head of 10 technician is rs . 1200 and the average salary / head of the rest is rs . 820 , the total no . of workers in the work - shop is ?
let the total number of workers be y . so sum of salary for all workers = sum of salary of 10 technician + sum of salary for other y - 10 workers . 10 x 1200 + 820 ( y - 10 ) = 1000 y ⇒ 12000 + 820 y - 8200 = 1000 y ⇒ 180 y = 3800 ∴ y = 190 / 9 so total number of workers = 190 / 9 d
a = 10 * 1200 b = 820 * 10 c = a - b d = 1000 - 820 e = c / d
a ) 91 , b ) 40 , c ) 80 , d ) 90 , e ) none of these
b
gcd(1080, 920)
the maximum number of student amoung them 1080 pens and 920 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :
"solution required number of student = h . c . f of 1080 and 920 = 40 . answer b"
a = math.gcd(1080, 920)
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
b
add(7, subtract(76, 77))
76 ^ 77 / 7 remainder is ?
as , 4 ^ 3 / 3 = 64 / 3 remainder = 1 and also 4 / 3 = 1 and 5 ^ 3 / 3 remainder = 2 and also 5 / 3 = 2 so it does not matter on the power value so the answer will be 76 / 7 remainder = 6 answer : b
a = 76 - 77 b = 7 + a
a ) 6 ^ 5 , b ) ( 15 / 4 ) ^ 5 , c ) 3 ^ 6 , d ) 6 ^ 3 , e ) 15 ^ 3
b
divide(power(30, 10), power(30, 5))
30 ^ 10 / 240 ^ 5 = ?
"30 ^ 10 / 240 ^ 5 = ? a . 6 ^ 5 b . ( 15 / 4 ) ^ 5 c . 3 ^ 6 d . 6 ^ 3 e . 15 ^ 3 - > 30 ^ 10 / 240 ^ 5 = ( 30 ^ 10 ) / ( 8 ^ 5 ) ( 30 ^ 5 ) = ( 30 ^ 5 ) / ( 8 ^ 5 ) = ( 2 ^ 5 ) ( 15 ^ 5 ) / ( 2 ^ 5 ) ( 4 ^ 5 ) = ( 15 / 4 ) ^ 5 . thus , b is the answer ."
a = 30 ** 10 b = 30 ** 5 c = a / b
a ) 380 , b ) 470 , c ) 560 , d ) 650 , e ) 740
a
multiply(20, subtract(20, const_1))
20 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ?
"2 * 20 c 2 = 2 * 190 = 380 the answer is a ."
a = 20 - 1 b = 20 * a
a ) 20 , b ) 24 , c ) 26 , d ) 28 , e ) 30
e
inverse(subtract(inverse(10), inverse(15)))
working alone , a can complete a certain kind of job in 15 hours . a and d , working together at their respective rates , can complete one of these jobs in 10 hours . in how many hours can d , working alone , complete one of these jobs ?
"let total time taken by d to complete the job = d total time taken by a to complete the job = 15 work done by a in an hour 1 / a = 1 / 15 working together a and d can complete the job in 10 hours 1 / a + 1 / d = 1 / 10 = > 1 / d = 1 / 10 - 1 / 15 = 1 / 10 - 1 / 15 = 1 / 30 = > d = 30 hours answer e"
a = 1/(10) b = 1/(15) c = a - b d = 1/(c)
a ) 15 days , b ) 13 days , c ) 22 days , d ) 11 days , e ) 19 days
a
add(divide(const_1, 12), divide(const_1, 20))
a can do a job in 12 days and b can do it in 20 days . a and b working together will finish twice the amount of work in - - - - - - - days ?
"a 1 / 12 + 1 / 20 = 2 / 15 15 / 2 * 2 = 15 days"
a = 1 / 12 b = 1 / 20 c = a + b
a ) 10 kg , b ) 12 kg , c ) 15 kg , d ) 16 kg , e ) 20 kg
a
multiply(divide(divide(multiply(subtract(const_100, 80), 40), const_100), subtract(const_100, 20)), const_100)
fresh grapes contain 80 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 40 kg of fresh grapes ?
"from the question we know : 40 kg * 80 % = 32 kg of water in the fresh grapes 40 kg - 32 kg of water = 8 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 20 % of the weight of the dry graps is water and we know that 8 kg is non - water mass we can set up the following equation : x = 1 / 5 ( x ) + 8 kg 4 / 5 ( x ) = 8 kg x = 10 kg answer - a"
a = 100 - 80 b = a * 40 c = b / 100 d = 100 - 20 e = c / d f = e * 100
a ) 7,858 , b ) 8,301 , c ) 14,667 , d ) 63,840 , e ) 146,667
e
divide(divide(multiply(55000, subtract(4.25, 3.85)), subtract(4.8, 4.65)), multiply(const_10, const_1000))
java house charges $ 4.25 for a cup of coffee that costs a total of $ 3.85 to make . cup ' o ' joe charges $ 4.80 for a cup that costs $ 4.65 to make . if java house sells 55000 cups of coffee , how many must cup ' o ' joe sell to make at least as much in total gross profit as its competitor does ?
java : profit / cup = 4.25 - 3.85 = 0.4 : no of cups = 55,000 : gross profit = 55,000 * 0.4 = 22,000 joe : profit / cup = 0.15 : gross profit = 22,000 : no of cups = 22,000 / 0.15 = 220,000 / 1.5 ( only closes is 146,667 ) answer e
a = 4 - 25 b = 55000 * a c = 4 - 8 d = b / c e = 10 * 1000 f = d / e
a ) 30 , b ) 35 , c ) 20 , d ) 18 , e ) 10
b
subtract(subtract(add(subtract(90, 11), 14), 44), 14)
in a neighborhood having 90 households , 11 did not have either a car or a bike . if 14 households had a both a car and a bike and 44 had a car , how many had bike only ?
"{ total } = { car } + { bike } - { both } + { neither } - - > 90 = 44 + { bike } - 14 + 11 - - > { bike } = 49 - - > # those who have bike only is { bike } - { both } = 49 - 14 = 35 . answer : b ."
a = 90 - 11 b = a + 14 c = b - 44 d = c - 14
a ) 2828 , b ) 2850 , c ) 1365 , d ) 1574 , e ) none of these
a
divide(62216, 22)
if the product of two numbers is 62216 and their h . c . f . is 22 , find their l . c . m .
"explanation : hcf * lcm = 62216 , because we know product of two numbers = product of hcf and lcm lcm = 62216 / 22 = 2828 option a"
a = 62216 / 22
a ) 80 % , b ) 105 % , c ) 120 % , d ) 96 % , e ) 138 %
d
multiply(divide(multiply(12, subtract(const_1, divide(20, const_100))), 10), const_100)
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 12 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ?
"x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,12 = 12 / 100 x = 80 * 12 / 100 x = 9.6 9.6 / 10 = 0.96 = 96 % , answer d"
a = 20 / 100 b = 1 - a c = 12 * b d = c / 10 e = d * 100
a ) 350 m , b ) 400 m , c ) 272 m , d ) 330 m , e ) 267 m
d
divide(subtract(1200, multiply(divide(45, 15), 180)), const_2)
a train crosses a tunnel of 1200 m in 45 sec , same train crosses another platform of length 180 m in 15 sec . then find the length of the train ?
length of the train be ‘ x ’ ( x + 1200 ) / 45 = ( x + 180 ) / 15 x + 1200 = 3 x + 540 2 x = 660 x = 330 m answer : d
a = 45 / 15 b = a * 180 c = 1200 - b d = c / 2
a ) 10 days , b ) 11 days , c ) 13 days , d ) 15 days , e ) 24 days
e
inverse(add(divide(6, multiply(8, 40)), divide(11, multiply(12, 40))))
if 8 men or 12 women can do a piece of work in 40 days , in how many days can the same work be done by 6 men and 11 women ?
"8 men = 12 women ( i . e 2 men = 3 women ) 12 women 1 day work = 1 / 40 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 40 = 1 / 480 so , 20 women work = 20 / 480 = 1 / 24 ans : 24 days answer : e"
a = 8 * 40 b = 6 / a c = 12 * 40 d = 11 / c e = b + d f = 1/(e)
a ) 42.25 cm 2 , b ) 57.5 cm 2 , c ) 42.5 cm 2 , d ) 47.5 cm 2 , e ) 52.5 cm 2
d
divide(multiply(19, 5), const_2)
if the sides of a triangle are 21 cm , 19 cm and 5 cm , what is its area ?
"the triangle with sides 21 cm , 19 cm and 5 cm is right angled , where the hypotenuse is 21 cm . area of the triangle = 1 / 2 * 19 * 5 = 47.5 cm 2 answer : d"
a = 19 * 5 b = a / 2
a ) 18 min , b ) 4 min , c ) 8 min , d ) 6 min , e ) 5 min
b
multiply(const_60, divide(subtract(90, 84), 90))
excluding stoppages , the speed of a bus is 90 km / hr and including stoppages , it is 84 km / hr . for how many minutes does the bus stop per hour ?
"due to stoppages , it covers 6 km less . time taken to cover 6 km = 6 / 90 * 60 = 4 min . answer : b"
a = 90 - 84 b = a / 90 c = const_60 * b
a ) 7 / 36 , b ) 9 / 35 , c ) 4 / 37 , d ) 8 / 35 , e ) 5 / 3
a
multiply(2, add(divide(const_1, 18), divide(const_1, 24)))
two persons a and b can complete a piece of work in 18 days and 24 days respectively . if they work together , what part of the work will be completed in 2 days ?
"a ' s one day ' s work = 1 / 18 b ' s one day ' s work = 1 / 24 ( a + b ) ' s one day ' s work = 1 / 18 + 1 / 24 = 7 / 72 the part of the work completed in 3 days = 2 ( 7 / 72 ) = 7 / 36 . answer : a"
a = 1 / 18 b = 1 / 24 c = a + b d = 2 * c
a ) 0.21 % , b ) 21 % , c ) 0.021 % , d ) 2.1 % , e ) none of these
d
multiply(divide(subtract(32, 11), 1000), const_100)
present birth rate is 32 per 1000 while death rate is 11 per 1000 . what will be the percentage increase in the rate of population ?
birth rate = 32 / 1000 = 0.032 death rate = 11 / 1000 = 0.011 increase in population = 0.032 - 0.011 = 0.021 = 2.1 % answer : d
a = 32 - 11 b = a / 1000 c = b * 100
a ) s . 45 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 120
e
divide(720, subtract(11, 5))
on selling 11 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is :
"( c . p . of 11 balls ) - ( s . p . of 11 balls ) = ( c . p . of 5 balls ) c . p . of 6 balls = s . p . of 11 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 6 = rs . 120 . answer : option e"
a = 11 - 5 b = 720 / a
a ) ( a ) 7 / 16 , b ) ( b ) 7 / 11 , c ) ( c ) 10 / 21 , d ) ( d ) 17 / 35 , e ) ( e ) 1 / 2
b
divide(multiply(subtract(const_10, multiply(divide(2, 5), const_10)), multiply(divide(7, 6), multiply(divide(2, 5), const_10))), add(multiply(multiply(divide(2, 5), const_10), multiply(divide(2, 5), const_10)), multiply(subtract(const_10, multiply(divide(2, 5), const_10)), multiply(divide(7, 6), multiply(divide(2, 5), const_10)))))
a lemonade stand sold only small and large cups of lemonade on tuesday . 2 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ?
"a simpler way i guess would be to think that in total 5 cups were sold . out of which 2 are small and 3 are large . now let the small ones cost $ 6 . so the large ones would cost $ 7 . so , 2 * 6 = 12 and 3 * 7 = 21 . total revenue was 12 + 21 = 33 and large cup sales as found above is 21 therefore answer is 21 / 33 i . e 7 / 11 b"
a = 2 / 5 b = a * 10 c = 10 - b d = 7 / 6 e = 2 / 5 f = e * 10 g = d * f h = c * g i = 2 / 5 j = i * 10 k = 2 / 5 l = k * 10 m = j * l n = 2 / 5 o = n * 10 p = 10 - o q = 7 / 6 r = 2 / 5 s = r * 10 t = q * s u = p * t v = m + u w = h / v
a ) 1.2 km , b ) 1.3 km , c ) 1.4 km , d ) 1.5 km , e ) 1.6 km
d
divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 6))
a boy is travelling from his home to school at 3 km / hr and reached 7 min late . next day he traveled at 6 km / hr and reached 8 min early . distance between home and school ?
"let the distance be x t 1 = x / 3 hr t 2 = x / 6 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 3 - x / 6 = 1 / 4 x / 6 = 1 / 4 x = 1.5 km answer is d"
a = 7 / const_60 b = 8 / const_60 c = a + b d = 1 / 6 e = c / d
a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 420 , e ) $ 260
d
subtract(multiply(add(multiply(15, add(const_3, const_2)), const_2), 15), multiply(30, const_12))
a parking garage rents parking spaces for $ 15 per week or $ 30 per month . how much does a person save in a year by renting by the month rather than by the week ?
"10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 15 = 780 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 30 = 360 $ 780 - 360 = 420 ans d"
a = 3 + 2 b = 15 * a c = b + 2 d = c * 15 e = 30 * 12 f = d - e
a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014
a
add(2001, divide(add(divide(15, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), subtract(6.30, 4.20))))
the price of commodity x increases by 40 cents every year , while the price of commodity y increases by 15 cents every year . in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 6.30 . in which year will the price of commodity x be 15 cents more than the price of commodity y ?
"the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 15 cents more than commodity y . $ 2.25 / 25 cents = 9 years the answer is 2001 + 9 years = 2010 . the answer is a ."
a = 15 / 100 b = 6 - 30 c = a + b d = 40 / 100 e = 6 - 30 f = d - e g = c / f h = 2001 + g
a ) 60 mph , b ) 72 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph
b
add(divide(add(multiply(80, 3), multiply(60, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33))
steve traveled the first 2 hours of his journey at 60 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ?
"distance traveled in 2 hours = 2 * 60 = 120 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 120 = 360 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 360 / 5 = 72 mph answer : b"
a = 80 * 3 b = 60 * 2 c = a + b d = 3 + 2 e = c / d f = 100 / 3 g = f - const_0_33 h = e + g
a ) rs . 83.33 , b ) rs . 110 , c ) rs . 112 , d ) rs . 120 , e ) rs . 140
d
multiply(divide(const_100, 10), 12)
a 12 % stock yielding 10 % is quoted at :
income of rs 10 on investment of rs 100 income of rs 12 on investment of ? = ( 12 * 100 ) / 10 = 120 answer : d
a = 100 / 10 b = a * 12
a ) 100.11 , b ) 10.98 , c ) 18.07 , d ) 45.01 , e ) 35.09
c
inverse(subtract(divide(subtract(const_1, multiply(inverse(100), const_2)), subtract(subtract(multiply(12, const_2), 7), const_2)), inverse(100)))
machine a takes 100 hours to complete a certain job and starts that job at 7 am . after two hour of working alone , machine a is joined by machine b and together they complete the job at 12 pm . how long would it have taken machine b to complete the jobif it had worked alone for the entire job ?
let us assume total job = 100 units a finishes 100 units in 100 hrs ( given ) hence a ( working rate ) = 1 units / hr now given that a works for 2 hr ( so 2 units done ) then a and b finish total work in 15 hours . hence a and b finish 98 units in 15 hours . of these 1 x 15 = 15 units were done by a . hence b did 83 units in 15 hours . hence b ( working rate ) = 83 / 15 units / hr hence b takes 100 x 15 / 83 = 18.07 hours to complete the job . answer c .
a = 1/(100) b = a * 2 c = 1 - b d = 12 * 2 e = d - 7 f = e - 2 g = c / f h = 1/(100) i = g - h j = 1/(i)
a ) 10 ^ 7 , b ) 10 ^ 8 , c ) 10 ^ 9 , d ) 10 ^ 10 , e ) 10 ^ 11
b
multiply(multiply(79, 48), add(add(multiply(multiply(const_0_25, const_1000), const_100), multiply(add(const_3, const_4), const_10)), const_3))
79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,070 bacteria each . how many bacteria are there approximately ?
"79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,070 bacteria each . how many bacteria are there approximately ? a . 10 ^ 7 b . 10 ^ 8 c . 10 ^ 9 d . 10 ^ 10 e . 10 ^ 11 - > due to approximately , 79 = 80 , 48 = 50 , 25,070 = 25,000 are derived , which makes ( 79 ) ( 48 ) ( 25,075 ) = ( 80 ) ( 50 ) ( 25,000 ) = 10 ^ 8 . the answer is b ."
a = 79 * 48 b = const_0_25 * 1000 c = b * 100 d = 3 + 4 e = d * 10 f = c + e g = f + 3 h = a * g
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500
a
divide(multiply(20, add(const_4, const_1)), const_2)
to fill a tank , 20 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to one - fifth of its present ?
"let the capacity of 1 bucket = x . then , the capacity of tank = 20 x . new capacity of bucket = 1 / 5 x therefore , required number of buckets = ( 20 x ) / ( 1 x / 5 ) = ( 20 x ) x 5 / 1 x = 100 / 1 = 100 answer is a ."
a = 4 + 1 b = 20 * a c = b / 2
a ) rs . 14,000 , b ) rs . 12,000 , c ) rs . 30,000 , d ) rs . 40,000 , e ) rs . 45,000
b
subtract(multiply(5, const_4), const_12)
a salesman â € ™ 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 ?
"explanation : [ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 12000 answer is b"
a = 5 * 4 b = a - 12
a ) 0.98 % , b ) 4.98 % , c ) 3.95 % , d ) 6.98 % , e ) 7.98 %
a
subtract(multiply(multiply(add(const_100, 32), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100)
shopkeeper rise price by 32 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ?
"let d initial price be 100 32 % rise now price = 132 / 100 * 100 = 132 10 % discount then price = 132 * 90 / 100 = 118.8 15 % discount then price = 118.8 * 85 / 100 = 100.98 so gain = 100.98 - 100 = 0.98 gain % = gain * 100 / cp = = > 0.98 * 100 / 100 = 0.98 % answer : a"
a = 100 + 32 b = 100 - 10 c = b / 100 d = a * c e = 100 - 15 f = e / 100 g = d * f h = g - 100
a ) rs . 3600 , b ) rs . 4000 , c ) rs . 3639 , d ) rs . 3632 , e ) rs . 3602
b
subtract(20000, multiply(divide(4, 5), 20000))
income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 20000 , then find his savings ?
"let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 20000 = > x = 4000 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 4000 answer : b"
a = 4 / 5 b = a * 20000 c = 20000 - b
a ) 34 % , b ) 24 % , c ) 22 % , d ) 38 % , e ) 8.5 %
d
multiply(divide(subtract(multiply(const_100, divide(17, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100)
one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 17 percent sugar by weight . the second solution was what percent sugar by weight ?
"instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 10 % + 1 part of x ( unknown ) % = 4 parts of 17 % = > x % = 68 % - 30 % = 38 % ans d it is ."
a = 17 / 100 b = 100 * a c = 1 / 4 d = c * 100 e = 100 - d f = 10 / 100 g = e * f h = b - g i = 1 / 4 j = i * 100 k = h / j l = k * 100
a ) 21.5 , b ) 22 , c ) 15.8 , d ) 12.8 , e ) 25
a
divide(subtract(multiply(8, 1.6), multiply(3, 1.4)), 1.3)
8 x 1.6 - 3 x 1.4 / 1.3 = ?
"given expression = ( 12.8 - 4.2 ) / 0.4 = 8.6 / 0.4 = 8.6 / 0.4 = 21.5 answer is a ."
a = 8 * 1 b = 3 * 1 c = a - b d = c / 1
a ) 0.5 l , b ) 1.5 l , c ) 2.5 l , d ) 3.5 l , e ) 4.5 l
c
subtract(3.5, const_1)
a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 3.5 t seconds , then what is the length of the platform ?
"the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 3.5 t ( l + p ) / ( l / t ) = 3.5 t p = 2.5 l the answer is c ."
a = 3 - 5
a ) a ) 1040 , b ) b ) 1045 , c ) c ) 1055 , d ) d ) 1190 , e ) e ) 1075
d
add(multiply(8, 70), multiply(9, 70))
tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 70 per kg . how much amount did he pay to the shopkeeper ?
"cost of 8 kg apples = 70 × 8 = 560 . cost of 9 kg of mangoes = 70 × 9 = 630 . total cost he has to pay = 560 + 630 = 1190 . d )"
a = 8 * 70 b = 9 * 70 c = a + b
a ) 70 % , b ) 45 % , c ) 48 % , d ) 30 % , e ) 40 %
b
multiply(divide(subtract(1220, 840), 840), const_100)
a cycle is bought for rs . 840 and sold for rs . 1220 , find the gain percent ?
"explanation : 840 - - - - 380 100 - - - - ? = > 45 % answer : b"
a = 1220 - 840 b = a / 840 c = b * 100
a ) – 15 , b ) – 2 , c ) - 3 , d ) 6 , e ) 15
b
subtract(3, divide(25, add(2, 3)))
for all numbers a and b , the operationis defined by ab = ( a + 2 ) ( b – 3 ) . if 3 x = – 25 , then x =
( 3 + 2 ) ( x - 3 ) = - 25 . . x - 3 = - 5 . . x = - 2 b
a = 2 + 3 b = 25 / a c = 3 - b
['a ) 7', 'b ) 11', 'c ) 13', 'd ) 16', 'e ) 26']
c
multiply(3.3, divide(8, 2))
the ratio of length to width of a rectangular showroom window is 3.3 to 2 . if the width of the window is 8 feet , what is the approximate length of the display in feet ?
explanation : letting l be the length of the window , the proportion for the ratio of the length to the width can be expressed in the following equation : 3.3 / 2 = l / 8 26.4 = 2 l 13.2 = l answer : option c
a = 8 / 2 b = 3 * 3
a ) 14 , b ) 13 , c ) 15 , d ) 17 , e ) 16
d
divide(170, const_2)
if n is a positive integer and the product of all the integers from 1 to n , inclusive , is a multiple of 170 , what is the least possible value of n ?
17 , since n needs to be the largest of the prime factors . therefore , the answer is d .
a = 170 / 2
a ) 12 , b ) 15 , c ) 23 , d ) 7 , e ) 5
d
subtract(50, add(add(divide(lcm(lcm(8, 5), 10), 8), multiply(divide(lcm(lcm(8, 5), 10), 5), 2)), multiply(divide(lcm(lcm(8, 5), 10), 10), 7)))
in a jar there are balls in different colors : blue , red , green and yellow . the probability of drawing a blue ball is 1 / 8 . the probability of drawing a red ball is 2 / 5 . the probability of drawing a green ball is 7 / 10 . if a jar can not contain more than 50 balls , how many yellow balls are in the jar ?
answer of 1 st just add the given probabilities p ( blue ) + p ( red ) + p ( green ) i . e 1 / 8 + 2 / 5 + 3 / 10 = 5 + 16 + 12 / 40 now we know p ( blue ) + p ( red ) + p ( green ) + p ( yellow ) = 1 33 / 40 + p ( yellow ) = 1 p ( yellow ) = 33 / 40 i . e why yellow balls are 7 . d
a = math.lcm(8, 5) b = math.lcm(a, 10) c = b / 8 d = math.lcm(8, 5) e = math.lcm(d, 10) f = e / 5 g = f * 2 h = c + g i = math.lcm(8, 5) j = math.lcm(i, 10) k = j / 10 l = k * 7 m = h + l n = 50 - m
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60
e
divide(multiply(subtract(4, 2.5), 100), 2.5)
a wholesaler wishes to sell 100 pounds of mixed nuts at $ 2.50 a pound . she mixes peanuts worth $ 3.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ?
from the question stem we know that we need a mixture of 100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y , we get x + y = 100 . since the wholesaler wants to sell the mixture of 100 pounds @ $ 2.50 , we can write this as : $ 2.5 * ( x + y ) = $ 1.5 x + $ 4 y from the equation x + y = 100 , we can rewrite y as y = 100 - x and substitute this into our equation to get : $ 2.5 * ( x + 100 - x ) = $ 1.5 x + $ 4 ( 100 - x ) if you solve for x , you will get x = 60 , and therefore y = 40 . so the wholesaler must use 40 pounds of cashews . you can substitute into the original equation to see that : $ 250 = $ 1.5 ( 60 ) + $ 4 ( 40 ) answer is e
a = 4 - 2 b = a * 100 c = b / 2
a ) 26 , b ) 30 , c ) 32 , d ) 34 , e ) 36
b
multiply(90, divide(2, 5))
of the 90 people in a room , 2 / 5 are women . if 2 / 3 of the people are married , what is the maximum number of women in the room who could be unmarried ?
"women = 2 / 5 * 90 = 36 married = 2 / 3 * 90 = 60 unmarried = 30 max ( un - married women ) = 30 b"
a = 2 / 5 b = 90 * a
a ) 16 hours , b ) 18 hours , c ) 20 hours , d ) 24 hours , e ) 28 hours
d
add(divide(5, add(9, 1)), divide(5, subtract(9, 1)))
speed of a boat in standing water is 9 kmph and the speed of the stream is 1 . 5 kmph . a man rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is
"upstream = 9 kmph - 1.5 kmph = 7.5 kmph downstream = 9 kmph + 1.5 kmph = 10.5 kmph total time taken = 105 / 7.5 + 105 / 10.5 = 24 hrs answer : d"
a = 9 + 1 b = 5 / a c = 9 - 1 d = 5 / c e = b + d
a ) 7000 , b ) 8300 , c ) 6000 , d ) 5000 , e ) 4000
b
divide(83, divide(subtract(7, 6), const_100))
in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 83 more candidates got selected than a . what was the number of candidates appeared from each state ?
"state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 83 more candidates got selected than state a from these , it is clear that 1 % of the total appeared candidates in state b = 83 = > total appeared candidates in state b = 83 x 100 = 8300 = > total appeared candidates in state a = total appeared candidates in state b = 8300"
a = 7 - 6 b = a / 100 c = 83 / b
a ) 16 hours , b ) 18 hours , c ) 20 hours , d ) 24 hours , e ) 26 hours
d
add(divide(105, add(9, 1.5)), divide(105, subtract(9, 1.5)))
speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph . a man rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is :
"speed of upstream = 9 - 1.5 = 7.5 kmph speed of down stream = 9 + 1.5 = 10.5 kmph total time = ( 105 / 7.5 ) + ( 105 / 10.5 ) = 24 hrs answer : d"
a = 9 + 1 b = 105 / a c = 9 - 1 d = 105 / c e = b + d
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16
e
add(add(4, 3), add(add(4, 4), 1))
for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 902 , what is the maximum possible sum of the length of x and the length of y ?
"we know that : x > 1 , y > 1 , and x + 3 y < 902 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 1000 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 902 so , 3 y < 390 = = > y < 130 so , y can take any value which is less than 130 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . e"
a = 4 + 3 b = 4 + 4 c = b + 1 d = a + c
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27
a
subtract(add(multiply(10, const_2), 5), 6)
alice is now 10 years older than bob . if in 6 years alice will be twice as old as bob , how old will alice be in 5 years ?
a = b + 10 so b = a - 10 . a + 6 = 2 ( b + 6 ) . a + 6 = 2 ( a - 10 + 6 ) . a + 6 = 2 a - 8 . a = 14 . in 5 years , alice will be 19 years old . the answer is a .
a = 10 * 2 b = a + 5 c = b - 6
a ) 30 % , b ) 35 % , c ) 25 % , d ) 40 % , e ) 50 %
c
divide(multiply(2,800, const_100), add(add(multiply(const_2, const_1000), const_100), 2,800))
in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 2,800 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 8,400 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?
"ratio of 2 bedroom apartment : 1 bedroom apartment = 2800 : 8400 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 4 ) * 100 - - - > 25 % answer : c"
a = 2 * 800 b = 2 * 1000 c = b + 100 d = c + 2 e = a / d
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
c
subtract(divide(add(180, 20), const_4), 20)
thabo owns exactly 180 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books does thabo own ?
"i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fiction books is 0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f 2 x + 40 0 nf x + 20 x total 3 x + 60 x 180 3 x + 60 + x = 180 x = 30 answer ( c . )"
a = 180 + 20 b = a / 4 c = b - 20
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8
c
divide(add(add(7, 6), 5), 3)
a school has 7 maths 6 physics and 5 chemistry teachers each teacher can teach 3 subjects max what is he minimum number of teachers required
total subjects = 7 + 6 + 5 = 18 max subjects by 1 teacher = 3 so , min of teachers required = 18 / 3 = 6 answer : c
a = 7 + 6 b = a + 5 c = b / 3
a ) 6350 , b ) 7357 , c ) 6328 , d ) 6750 , e ) 7560
d
multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 10))), rectangle_area(75, 55)), 10)
a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 10 per sq m ?
"area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 10 = rs . 6750 answer : d"
a = 2 * 5 b = 75 + a c = 2 * 5 d = 55 + c e = rectangle_area - ( f = e * rectangle_area
a ) 0.33 , b ) 0.44 , c ) 0.48 , d ) 44 , e ) 48
e
multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.36, 100))
in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.36 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]
"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 36 cents ; a half - liter of kerosene costs as much as 8 eggs - - > 8 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 16 eggs = 16 / 12 * 36 = 48 cents . answer : e ."
a = 1 / 2 b = 8 / a c = b / 12 d = 0 * 36 e = c * d
a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 4
d
divide(4, add(8, 4))
at a certain high school , the senior class is twice the size of the junior class . if 3 / 8 of the seniors and 1 / 4 of the juniors study japanese , what fraction of the students in both classes study japanese ?
"start by deciding on a number of students to represent the number of students in the senior class . for this example i will choose 200 students . that would make the number of students in the junior class 100 . then we can find out how many students are taking japanese in each grade and add them together . ( 3 / 8 ) * 200 = 75 and ( 1 / 4 ) * 100 = 25 . 75 + 25 = 100 . there are a total of 300 students in the junior class and senior class combined ( 100 + 200 = 300 ) , and there are 100 total students in japanese , so 100 students in japanese / 300 total students equals 1 / 3 of the students in both classes that study japanese . answer : d"
a = 8 + 4 b = 4 / a
a ) 150 , b ) 151 , c ) 250 , d ) 350 , e ) none
b
subtract(multiply(12, 604), multiply(add(11, divide(9, 12)), 604))
in a school with 604 students , the average age of the boys is 12 years and that of the girls is 11 years . if the average age of the school is 11 years 9 months , then the number of girls in the school is
"sol . let the number of grils be x . then , number of boys = ( 600 - x ) . then , ( 11 3 / 4 × 604 ) ⇔ 11 x + 12 ( 604 - x ) ⇔ x = 7248 - 7097 ⇔ 151 . answer b"
a = 12 * 604 b = 9 / 12 c = 11 + b d = c * 604 e = a - d
a ) 44 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease
a
subtract(divide(multiply(add(80, const_100), subtract(const_100, 20)), const_100), const_100)
when the price of an article was reduced by 20 % its sale increased by 80 % . what was the net effect on the sale ?
"if n items are sold for $ p each , revenue is $ np . if we reduce the price by 20 % , the new price is 0.8 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.8 p ) ( 1.8 n ) = 1.44 np , which is 1.44 times the old revenue , so is 44 % greater . answer : a"
a = 80 + 100 b = 100 - 20 c = a * b d = c / 100 e = d - 100
a ) 35 , b ) 20 , c ) 47 , d ) 36 , e ) 31
e
subtract(110, divide(110, add(divide(2, 5), const_1)))
a 110 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ?
"1 : 2 / 5 = 5 : 2 2 / 7 * 110 = 31 answer : e"
a = 2 / 5 b = a + 1 c = 110 / b d = 110 - c
a ) 24.58 , b ) 24.5 , c ) 26 , d ) 27 , e ) 28
e
multiply(divide(8, subtract(9, 7)), 7)
sachin is younger than rahul by 8 years . if the ratio of their ages is 7 : 9 , find the age of sachin
"if rahul age is x , then sachin age is x - 8 , so ( x - 8 ) / x = 7 / 9 = > 9 x - 72 = 7 x = > 2 x = 72 = > x = 36 so sachin age is 36 - 8 = 28 answer : e"
a = 9 - 7 b = 8 / a c = b * 7
a ) 12 , b ) 12.1 , c ) 13 , d ) 14 , e ) 15
b
divide(add(110, 132), multiply(72, const_0_2778))
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length
"speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 20 = 12.1 sec . answer : option b"
a = 110 + 132 b = 72 * const_0_2778 c = a / b
a ) 44 % , b ) 40 % , c ) 50 % , d ) 56.25 % , e ) 36.25 %
d
multiply(subtract(divide(add(const_100, 25), subtract(const_100, 20)), const_1), const_100)
a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 25 % impurities to the product . find the net profit percentage of the dealer ?
"the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 25 % impurities to the product . it means that he is selling $ 5 worth of product for $ 6.25 . so his profit is $ 6.25 - $ 4 = $ 2.25 and his profit percent is ( 2.25 / 4 ) * 100 = 56.25 % answer : - d"
a = 100 + 25 b = 100 - 20 c = a / b d = c - 1 e = d * 100
a ) 198 mph , b ) 110 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph
d
divide(add(110, 90), const_2)
i flew my tiny seaplane to visit my mother . on the flight up , i flew at 110 mph . on the way home , i flew 90 mph . what was my average speed for the trip ?
( 110 mph + 90 mph ) / 2 = 100 mph correct answer is : d
a = 110 + 90 b = a / 2
a ) 85 yrs , b ) 95 yrs , c ) 93 yrs , d ) 82 yrs , e ) 90 yrs
b
add(add(multiply(const_2, 30), 30), 5)
in 30 years , a will be twice as old as b was 30 years ago . if a is now 5 years older than b , the present age of b is ?
"let b ' s present age = x years then a ' s present age = x + 5 years x + 5 + 30 = 2 ( x - 30 ) x + 35 = 2 x - 60 x = 95 years answer is b"
a = 2 * 30 b = a + 30 c = b + 5
a ) 3.6 sec , b ) 40 sec , c ) 36 sec , d ) 72 sec , e ) none of these
b
multiply(multiply(divide(divide(add(280, 120), const_1000), subtract(45, 9)), const_60), const_60)
a jogger running at 9 kmph along side a railway track is 280 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = ( 45 – 9 ) km / h = 36 km / h = ( 36 × 5 ⁄ 18 ) m / sec = 10 m / sec distance to be covered = ( 280 + 120 ) m = 400 m . ∴ time taken = ( 400 ⁄ 10 ) sec = 40 sec . answer b"
a = 280 + 120 b = a / 1000 c = 45 - 9 d = b / c e = d * const_60 f = e * const_60
a ) 5 sec , b ) 25 sec , c ) 20 sec , d ) 15 sec , e ) 10 sec
c
divide(180, add(7, 2))
an escalator moves towards the top level at the rate of 7 ft . sec and its length is 180 feet . if a person walks on the moving escalator at the rate of 2 feet per second towards the top level , how much time does he take to cover the entire length .
"explanation : time taken to cover the entire length = tot . dist / resultant speed = 180 / ( 7 + 2 ) = 20 sec answer : c"
a = 7 + 2 b = 180 / a
a ) 1077 m , b ) 804 m , c ) 978 m , d ) 1048 m , e ) 878 v
d
divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 1000), const_100)
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 1000 resolutions ?
"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 1000 resolutions . = 1000 * 2 * 22 / 7 * 22.4 = 104800 cm = 1408 m answer : d"
a = 3 + 4 b = a * 3 c = b + 1 d = 3 + 4 e = c / d f = e * 22 g = f * 2 h = g * 1000 i = h / 100
a ) 1500 , b ) 1893 , c ) 1979 , d ) 1900 , e ) 1278
a
subtract(divide(subtract(multiply(2000, 60), multiply(2000, 15)), 20), 2000)
a garrison of 2000 men has provisions for 60 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ?
"2000 - - - - 60 2000 - - - - 45 x - - - - - 20 x * 20 = 2000 * 45 x = 4500 2000 - - - - - - - 1500 answer : a"
a = 2000 * 60 b = 2000 * 15 c = a - b d = c / 20 e = d - 2000
['a ) 4225', 'b ) 4096', 'c ) 4356', 'd ) insufficient data', 'e ) none of these']
c
power(divide(add(131, 1), const_2), const_2)
anil grows tomatoes in his backyard which is in the shape of a square . each tomato takes 1 cm 2 in his backyard . this year , he has been able to grow 131 more tomatoes than last year . the shape of the backyard remained a square . how many tomatoes did anil produce this year ?
detailed solution let the area of backyard be x 2 this year and y 2 last year ∴ x 2 - y 2 = 131 = ) ( x + y ) ( x - y ) = 131 now , 131 is a prime number ( a unique one too . check out its properties on google ) . also , always identify the prime number given in a question . might be helpful in cracking the solution . = ) ( x + y ) ( x - y ) = 131 x 1 = ) x + y = 131 x - y = 1 = ) 2 x = 132 = ) x = 66 and y = 65 ∴ number of tomatoes produced this year = 662 = 4356 correct choice ( c )
a = 131 + 1 b = a / 2 c = b ** 2
a ) 2 , b ) 9 , c ) 3 , d ) 8 , e ) 6
d
subtract(multiply(1, const_12), divide(multiply(1, const_12), add(divide(1, 5), const_1)))
when asked what the time is , a person answered that the amount of time left is 1 / 5 of the time already completed . what is the time .
"a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 15 x ⇒ x = 2024 − x = 15 x ⇒ x = 20 time is 8 pm answer : d"
a = 1 * 12 b = 1 * 12 c = 1 / 5 d = c + 1 e = b / d f = a - e
a ) rs . 1200 , b ) rs . 1690 , c ) rs . 1600 , d ) rs . 1700 , e ) rs . 1500
d
subtract(1717, divide(multiply(subtract(1734, 1717), 1), 2))
a sum of money at simple interest amounts to rs . 1717 in 1 year and to rs . 1734 in 2 years . the sum is :
"s . i . for 1 year = rs . ( 1734 - 1717 ) = rs . 17 . principal = rs . ( 1717 - 17 ) = rs . 1700 . answer : option d"
a = 1734 - 1717 b = a * 1 c = b / 2 d = 1717 - c
a ) 21 , b ) 20 , c ) 23 , d ) 27 , e ) 31
c
divide(subtract(997, 8), 43)
if remainder is 8 , quotient is 43 and dividend is 997 then what is divisor ?
we know dividend = divisor * quotient + remainder = = = > 997 = divisor * 43 + 8 = = = = = > 989 / 43 = divisor = = = > divisor = 23 ans - c
a = 997 - 8 b = a / 43
a ) 334 , b ) 300 , c ) 376 , d ) 288 , e ) 271
b
divide(240, divide(80, const_100))
victor gets 80 % marks in examinations . if these are 240 marks , find the maximum marks .
"let the maximum marks be m then 80 % of m = 240 ⇒ 80 / 100 × m = 240 ⇒ m = ( 240 × 100 ) / 80 ⇒ m = 24000 / 80 ⇒ m = 300 therefore , maximum marks in the examinations are 300 answer : b"
a = 80 / 100 b = 240 / a
a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 105.00 , e ) none of these
d
divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(10800, multiply(3, 4)))), multiply(4, sqrt(divide(10800, multiply(3, 4))))), 25), const_100)
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10800 sq . m , the cost of fencing the field @ 25 paise per metre is
"solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 10800 ⇔ 12 x 2 = 10800 ⇔ x 2 = 900 ⇔ x = 30 . so , length = 90 m and breadth = 120 m . perimeter = [ 2 ( 90 + 120 ) ] m = 420 m . ∴ cost of fencing = rs . ( 0.25 × 420 ) = rs . 105.00 . answer d"
a = 3 * 4 b = 10800 / a c = math.sqrt(b) d = 3 * c e = 3 * 4 f = 10800 / e g = math.sqrt(f) h = 4 * g i = rectangle_perimeter * ( j = i / 25