options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 150 % , b ) 160 % , c ) 156 % , d ) 158 % , e ) 162 % | c | multiply(subtract(power(add(const_1, divide(60, const_100)), const_2), const_1), const_100) | what will be the percentage increase in the area of the cube ' s surface if each of the cube ' s edges grows by 60 % ? | "the question is very easy . my logic is the following : a surface = 6 * a ^ 2 after 60 % increase a surface = 6 * ( ( 1.6 a ) ^ 2 ) = 6 * 2.56 * a ^ 2 the increase in the surface area = ( 6 * 2.56 * a ^ 2 - 6 * a ^ 2 ) / 6 * a ^ 2 = ( 6 * a ^ 2 ( 2.56 - 1 ) ) / ( 6 * a ^ 2 ) = 2.56 - 1 = 1.56 = 156 % answer : c" | a = 60 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = d * 100
|
a ) 160 , b ) 220 , c ) 240 , d ) 360 , e ) 420 | c | add(divide(multiply(divide(55, const_100), 20), subtract(divide(60, const_100), divide(55, const_100))), 20) | the workforce of company x is 60 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 20 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 20 ) 0.05 x = 11 x = 220 - this is quantity of employees before adding 2020 men so after adding it will be 240 answer is c | a = 55 / 100
b = a * 20
c = 60 / 100
d = 55 / 100
e = c - d
f = b / e
g = f + 20
|
a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 20 | e | subtract(subtract(subtract(90, 10), add(90, 10)), 10) | if | 20 x - 10 | = 90 , then find the product of the values of x ? | "| 20 x - 10 | = 90 20 x - 10 = 90 or 20 x - 10 = - 90 20 x = 100 or 20 x = - 80 x = 5 or x = - 4 product = - 4 * 5 = - 20 answer is e" | a = 90 - 10
b = 90 + 10
c = a - b
d = c - 10
|
a ) 3 . , b ) 2 . , c ) 1 / 2 . , d ) 4 , e ) there is n ' t enough data to answer the question . | d | add(3, divide(multiply(3, 1), 3)) | two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores is 1 / 3 , what is the value of x / y ? | answer is d : 4 x - y = ( x + y ) / 2 solving for x / y = 4 | a = 3 * 1
b = a / 3
c = 3 + b
|
a ) 1 / 20 , b ) 1 / 15 , c ) 1 / 12 , d ) 1 / 8 , e ) 1 / 2 | a | multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2))) | let a be the event that a randomly selected two digit number is divisible by 2 and let b be the event that a randomly selected two digit number is divisible by 10 . what is p ( a and b ) ? | "p ( a and b ) = 1 / 2 * 1 / 10 = 1 / 20 the answer is a ." | a = 2 - 1
b = 2 - 1
c = b * 2
d = a / c
e = 2 - 1
f = e * 2
g = 2 - 1
h = g * 2
i = f / h
j = d * i
|
a ) 1 hour , b ) 2 hour , c ) 3 hour , d ) 4 hour , e ) 5 hour | a | divide(600, divide(multiply(200, 30), const_10)) | in a flight of 600 km , an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by 200 km / hr and the time of flight increased by 30 minutes . the duration of the flight is : | "let the duration of the flight be x hours then 600 / x - 600 / x ( 1 / 2 ) = 200 = 600 / x - 1200 / 2 x + 1 = 200 = x ( 2 x + 1 ) = 3 = 2 xsquare + x - 3 = 0 = ( 2 x + 3 ) ( x - 1 ) = 0 = x = 1 hr ( neglecting the - ve value of x ) answer : option a" | a = 200 * 30
b = a / 10
c = 600 / b
|
a ) 15 km , b ) 30 km , c ) 45 km , d ) 50 km , e ) 60 km | b | multiply(multiply(divide(divide(23, const_60), add(add(divide(const_1, 50), divide(const_1, 100)), divide(const_1, 120))), const_3), const_1000) | a person travels equal distances with speeds of 50 km / hr , 100 km / hr and 120 km / hr and takes a total time of 23 minutes . the total distance is ? | "let the total distance be 3 x km x / 50 + x / 100 + x / 120 = 23 / 60 23 x / 600 = 23 / 60 23 x = 230 x = 10 km total distance = 3 x = 30 km answer is b" | a = 23 / const_60
b = 1 / 50
c = 1 / 100
d = b + c
e = 1 / 120
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 7.5 hr , b ) 6.98 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | b | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 25))) | two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 25 hr . if all pipes are opened simultaneously , then the cistern will be filled in | "solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 β 1 / 25 = 1 / 7 hence , tank will be filled in 6.98 hour option ( b )" | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / 25
e = c - d
f = 1/(e)
|
a ) 223888 , b ) 143547 , c ) 2607778 , d ) 126997 , e ) 127811 | b | add(multiply(add(multiply(multiply(7, 2), const_100), multiply(7, 5)), const_100), subtract(add(multiply(7, 2), multiply(7, 5)), 2)) | if 5 + 3 + 2 = 151022 , 9 + 2 + 4 = 183652 , then 7 + 2 + 5 = ? | if the given number is a + b + c then a . b | a . c | a . b + a . c - b β β 5 + 3 + 2 = 5.3 | 5.2 | 5.3 + 5.2 - 3 = 151022 β β 9 + 2 + 4 = 9.2 | 9.4 | 9.2 + 9.4 - 2 = 183652 7 + 2 + 5 = 7.2 | 7.5 | 7.2 + 7.5 - 2 = 143547 answer : b | a = 7 * 2
b = a * 100
c = 7 * 5
d = b + c
e = d * 100
f = 7 * 2
g = 7 * 5
h = f + g
i = h - 2
j = e + i
|
['a ) 120 m', 'b ) 145 m', 'c ) 130 m', 'd ) 140 m', 'e ) 122 m'] | d | divide(divide(880, const_pi), const_2) | the inner circumference of a circle race track 18 m wide is 880 m . find the radius of the outer circle . | let inner radius be r metres then 2 Ο r = 640 2 Γ 22 / 7 r = 880 44 / 7 r = 880 r = 880 Γ 7 / 44 = 140 m answer is d . | a = 880 / math.pi
b = a / 2
|
a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 49 | e | multiply(multiply(4, divide(14, 4)), divide(14, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 14 mat - weavers in 14 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 14 : : 4 : x days 4 : 14 4 * 4 * x = 14 * 14 * 4 x = ( 14 * 14 * 4 ) / ( 4 x 4 ) x = 49 . answer is e ." | a = 14 / 4
b = 4 * a
c = 14 / 4
d = b * c
|
a ) 76 , b ) 5776 , c ) 304 , d ) 3600 , e ) none | d | power(multiply(4, 15), const_2) | find β ? / 15 = 4 ? | "answer let β n / 15 = 4 then β n = 15 x 4 = 60 β΄ n = 60 x 60 = 3600 . correct option : d" | a = 4 * 15
b = a ** 2
|
a ) β 48 , b ) β 1 , c ) 2 , d ) 46 , e ) 67 | b | subtract(subtract(subtract(subtract(add(add(1, 20), subtract(1, 20)), const_1), const_1), const_1), const_1) | if a ( a - 1 ) = 20 and b ( b - 1 ) = 20 , where a β b , then a + b = | "i . e . if a = 5 then b = - 4 or if a = - 4 then b = 5 but in each case a + b = - 5 + 4 = - 1 answer : option b" | a = 1 + 20
b = 1 - 20
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) a ) 3800 , b ) b ) 4200 , c ) c ) 4400 , d ) d ) 4500 , e ) e ) 4600 | d | floor(divide(3240, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100)))) | 10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 3240 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 80 / 100 ) = 3240 x = 4500 answer : d" | a = 100 - 10
b = a / 100
c = 100 - 20
d = c / 100
e = b * d
f = 3240 / e
g = math.floor(f)
|
a ) 1642 , b ) 1640 , c ) 1632 , d ) 1688 , e ) 6386 | d | subtract(2743, divide(multiply(multiply(3, 5), 2743), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2743 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2743 - x ) * 3 * 5 ) / 100 24 x / 100 = 41145 / 100 - 15 x / 100 39 x = 41145 = > x = 1055 second sum = 2743 β 1055 = 1688 answer : d" | a = 3 * 5
b = a * 2743
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2743 - f
|
a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min | c | multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10) | tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many t minutes it takes john to reach the station ? | since the question states β after 15 minutes β , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john β s speed is 10 km / hour , which means 1 km / 6 minutes . as he has to travel 5 / 4 km more , it is going to take him 6 ( 5 / 4 ) minutes . hence , t = 6 ( 5 / 4 ) = 15 / 2 minutes . the correct answer is c . | a = 15 - 10
b = 15 / const_60
c = a * b
d = c * const_60
e = d / 10
f = e / 2
g = f - const_0_25
h = g * 10
|
a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 225 | e | divide(18, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 18 . if n / j = 134.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 18 so , 134.08 = 134 + 18 / j so , j = 225 ans e" | a = 4 * 10
b = a + 2
c = 100 + b
d = 134 - 8
e = 18 / d
|
a ) 5 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these | c | divide(multiply(multiply(divide(const_1, 15), 10), 18), const_2) | a can finish a work in 18 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 18 * 1 / 3 ) = 6 days . correct option : c | a = 1 / 15
b = a * 10
c = b * 18
d = c / 2
|
a ) 240 , b ) 272 , c ) 295 , d ) 360 , e ) 252 | b | divide(add(1375, 15), subtract(6, const_1)) | the difference of two numbers is 1375 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1375 ) . x + 1375 = 6 x + 15 5 x = 1360 x = 272 smaller number = 270 . answer b" | a = 1375 + 15
b = 6 - 1
c = a / b
|
a ) 9400 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000 | a | subtract(subtract(7600, multiply(7600, divide(10, const_100))), multiply(subtract(7600, multiply(7600, divide(10, const_100))), divide(10, const_100))) | the population of a town is 7600 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ? | "formula : ( after = 100 denominator ago = 100 numerator ) 7600 Γ£ β 100 / 90 Γ£ β 100 / 90 = 9382 a )" | a = 10 / 100
b = 7600 * a
c = 7600 - b
d = 10 / 100
e = 7600 * d
f = 7600 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 22.2 , b ) 22.1 , c ) 15.3 , d ) 15.8 , e ) 14.6 | d | divide(add(add(add(add(add(add(add(add(add(const_3, add(add(const_1, const_3), const_1)), add(add(const_1, const_3), const_3)), add(10, const_1)), add(10, const_3)), add(10, add(add(const_1, const_3), const_3))), add(add(10, add(add(const_1, const_3), add(add(const_1, const_3), const_1))), add(const_1, const_3))), add(add(10, add(add(const_1, const_3), add(add(const_1, const_3), const_1))), 10)), add(add(add(add(add(const_1, const_3), add(add(const_1, const_3), const_1)), add(add(const_1, const_3), const_3)), add(10, const_1)), 10)), add(10, add(add(const_1, const_3), add(add(const_1, const_3), const_1)))), 10) | the average of first 10 prime numbers which are odd is ? | explanation : sum of first 10 prime no . which are odd = 158 average = 158 / 10 = 15.8 answer : d | a = 1 + 3
b = a + 1
c = 3 + b
d = 1 + 3
e = d + 3
f = c + e
g = 10 + 1
h = f + g
i = 10 + 3
j = h + i
k = 1 + 3
l = k + 3
m = 10 + l
n = j + m
o = 1 + 3
p = 1 + 3
q = p + 1
r = o + q
s = 10 + r
t = 1 + 3
u = s + t
v = n + u
w = 1 + 3
x = 1 + 3
y = x + 1
z = w + y
A = 10 + z
B = A + 10
C = v + B
D = 1 + 3
E = 1 + 3
F = E + 1
G = D + F
H = 1 + 3
I = H + 3
J = G + I
K = 10 + 1
L = J + K
M = L + 10
N = C + M
O = 1 + 3
P = 1 + 3
Q = P + 1
R = O + Q
S = 10 + R
T = N + S
U = T / 10
|
a ) 4 , b ) 5 , c ) 56 , d ) 28 , e ) 7 | d | subtract(add(const_100, 60), add(divide(multiply(add(const_100, 60), 20), const_100), const_100)) | on increasing the price of t . v . sets by 60 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ? | explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 Γ 100 ) = rs . 10000 new sale value = rs . ( 160 Γ 80 ) = rs . 12800 increase % = 2800 / 10000 Γ 100 = 28 % answer : d | a = 100 + 60
b = 100 + 60
c = b * 20
d = c / 100
e = d + 100
f = a - e
|
a ) 40 - 43 , b ) 39 - 42 , c ) 39 - 42 , d ) 38 - 41 , e ) 36 - 40 | e | add(divide(multiply(35.50, 2), const_100), 35.50) | a meal cost $ 35.50 adn there was no tax . if the tip was more than 2 pc but less than 12 pc of the price , then the total amount paid should be : | "2 % ( 35.5 ) = 0.71 12 % ( 35.5 ) = 4.26 total amount could have been 35.5 + 0.71 and 35.5 + 4.26 = > could have been between 36.21 and 39.76 = > approximately between 36 and 40 answer is e ." | a = 35 * 50
b = a / 100
c = b + 35
|
a ) 504 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1200) | in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1200 employees , then the no . of female employees who are computer literate ? | solution : total employees , = 1200 female employees , 60 % of 1200 . = ( 60 * 1200 ) / 100 = 720 . then male employees , = 480 50 % of male are computer literate , = 240 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1200 ) / 100 = 744 computer literate . thus , female computer literate = 744 - 240 = 504 . answer : option a | a = 62 / 100
b = 60 / 100
c = 1 - b
d = 50 / 100
e = c * d
f = a - e
g = f * 1200
|
a ) 9 , b ) 16 , c ) 21 , d ) 25 1 / 3 , e ) 28 1 / 2 | c | multiply(40, divide(multiply(3, 2), multiply(4, 3))) | a paint store mixes 3 / 4 pint of red paint and 2 / 3 pint of white paint to make a new paint color called perfect pink . how many pints of red paint would be needed to make 40 pints of perfect pink paint ? | "3 / 4 pint is required to make 3 / 4 + 2 / 3 = 17 / 12 pint of perfect pink so 17 / 12 pint requires 3 / 4 pint of red . . 1 pint will require 3 / 4 * 12 / 17 = 9 / 17 . . 40 pints will require 9 / 17 * 40 = 21 pints . . c" | a = 3 * 2
b = 4 * 3
c = a / b
d = 40 * c
|
a ) 80 / 8 , b ) 80 / 9 , c ) 80 / 5 , d ) 80 / 3 , e ) 80 / 1 | b | divide(100, multiply(add(72, 90), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 72 kmph and 90 kmph . in how much time will the trains cross each other ? | "relative speed = ( 72 + 90 ) * 5 / 18 = 9 * 5 = 45 mps . the time required = d / s = ( 100 + 100 + 200 ) / 45 = 400 / 45 = 80 / 9 sec . answer : b" | a = 72 + 90
b = a * const_0_2778
c = 100 / b
|
a ) rs . 11.81 , b ) rs . 19.93 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none | b | divide(multiply(13, add(const_100, 15)), subtract(const_100, 25)) | a fruit seller sells mangoes at the rate of rs . 13 per kg and thereby loses 25 % . at what price per kg , he should have sold them to make a profit of 15 % ? | "solution 75 : 13 = 115 : x x = ( 13 Γ£ β 115 / 75 ) = rs . 19.93 hence , s . p per kg = rs . 19.93 answer b" | a = 100 + 15
b = 13 * a
c = 100 - 25
d = b / c
|
a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 60 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 40))), 30)) | mahesh can do a piece of work in 40 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 60 days = 20 * 1 / 40 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 1 / 2 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 2 = 60 days answer is e" | a = 1 / 40
b = 20 * a
c = 1 - b
d = c / 30
e = 1 / d
|
a ) $ 2800 , b ) $ 2464 , c ) $ 2650 , d ) $ 2732 , e ) $ 2400 | e | multiply(add(const_1, divide(40, const_100)), original_price_before_gain(40, 2240)) | a store β s selling price of $ 2240 for a certain printer would yield a profit of 40 percent of the store β s cost for the printer . what selling price would yield a profit of 50 percent of the printer β s cost ? | "1.4 x = 2240 x = 2240 / 1.4 so , 1.5 x = 2240 * 1.5 / 1.4 = 2400 answer : - e" | a = 40 / 100
b = 1 + a
c = b * original_price_before_gain
|
a ) 25 , b ) 30 , c ) 35 , d ) 20 , e ) none of these | a | divide(multiply(50, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | a contractor undertakes to built a walls in 50 days . he employs 50 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | "50 men complete 0.4 work in 25 days . applying the work rule , m 1 Γ d 1 Γ w 2 = m 2 Γ d 2 Γ w 1 we have , 50 Γ 25 Γ 0.6 = m 2 Γ 25 Γ 0.4 or m 2 = 50 Γ 25 Γ 0.6 / 25 Γ 0.4 = 75 men answera" | a = 100 - 40
b = a / 100
c = 50 * b
d = 4 / 10
e = c / d
|
a ) 784596 , b ) 845796 , c ) 804670 , d ) 784596 , e ) 864520 | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 500), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 500 miles into meters ? | "1 mile = 1609.34 meters 500 mile = 500 * 1609.34 = 804670 meters answer is c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 500
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 16 , b ) 18 , c ) 19 , d ) 20 , e ) 22 | d | subtract(divide(subtract(multiply(12, 5), 20), const_2), const_2) | the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 12 , and k < m < r < s < t . if t is 20 , what is the greatest possible value of the median of the 5 integers ? | "we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 12 which means that in effect , all numbers are equal to 12 . if the largest number is 20 , it is 8 more than 12 . we need r to be maximum so k and m should be as small as possible to get the average of 12 . since all the numbers are positive integers , k and m can not be less than 1 and 2 respectively . 1 is 11 less than 12 and 2 is 10 less than 12 which means k and m combined are 21 less than the average . 20 is already 8 more than 12 and hence we only have 21 - 8 = 13 extra to distribute between r and s . since s must be greater than r , r can be 12 + 6 = 18 and s can be 12 + 7 = 19 . so r is 18 . answer ( d )" | a = 12 * 5
b = a - 20
c = b / 2
d = c - 2
|
a ) 12 days , b ) 3 days , c ) 4 days , d ) 5 days , e ) 6 days | a | inverse(add(inverse(36), multiply(const_2, inverse(36)))) | a work as fast as b . if b can complete a work in 36 days independently , the number of days in which a and b can together finish the work in ? | "ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 18 b ' s 1 day work = 1 / 36 a + b 1 day work = 1 / 18 + 1 / 36 = 3 / 36 = 1 / 12 a and b can finish the work in 12 days answer is a" | a = 1/(36)
b = 1/(36)
c = 2 * b
d = a + c
e = 1/(d)
|
a ) 3 : 4 , b ) 4 : 3 , c ) 4 : 5 , d ) 5 : 4 , e ) 8 : 7 | a | divide(multiply(const_60, 2), add(multiply(const_60, 2), 40)) | a car covers a distance of 200 km in 2 hours 40 minutes , whereas a jeep covers the same distance in 2 hours . what is the ratio of t | a 3 : 4 | a = const_60 * 2
b = const_60 * 2
c = b + 40
d = a / c
|
a ) 25 , b ) 83 , c ) 45 , d ) 53 , e ) 64 | a | subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(54, 7))) | the average of 13 numbers is 60 . average of the first 7 of them is 54 and that of the last 7 is 61 . find the 8 th number ? | "sum of all the 13 numbers = 13 * 60 = 780 sum of the first 7 of them = 7 * 54 = 378 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 378 - 780 = 25 . answer : a" | a = 8 * 61
b = 13 * 60
c = 54 * 7
d = b - c
e = a - d
|
a ) 8 years , b ) 2 years , c ) 7 years , d ) 8 years , e ) 5 years | d | divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 22 years older than his son . in three years , his age will be thrice the age of his son . the present age of this son is | "let ' s son age is x , then father age is x + 22 . = > 3 ( x + 3 ) = ( x + 22 + 3 ) = > 3 x + 9 = x + 25 = > 2 x = 16 years = 8 years answer : d" | a = 2 * 2
b = a - 2
c = 22 - b
d = 2 - 1
e = c / d
|
a ) $ 27 , b ) $ 29 , c ) $ 31 , d ) $ 33 , e ) $ 35 | e | subtract(multiply(1.7, 70), multiply(1.2, 70)) | for each color copy , print shop x charges $ 1.20 and print shop y charges $ 1.70 . how much greater is the charge for 70 color copies at print shop y than at print shop x ? | the difference in the two prices is $ 1.70 - $ 1.20 = $ 0.50 for each color copy . each color copy will cost an extra $ 0.50 at print shop y . 70 * $ 0.50 = $ 35 the answer is e . | a = 1 * 7
b = 1 * 2
c = a - b
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a ) 42 minutes , b ) 51 minutes , c ) 39 minutes , d ) 40 minutes 20 seconds , e ) none of these | b | multiply(divide(850, subtract(add(40, 30), 20)), const_3) | pipe a fills a tank of capacity 850 liters at the rate of 40 liters a minute . another pipe b fills the same tank at the rate of 30 liters a minute . a pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute . if pipe a is kept open for a minute and then closed and pipe b is open for a minute and then closed and then pipe c is open for a minute and then closed and the cycle is repeated , when will the tank be full ? | "in one cycle they fill 40 + 30 - 20 = 50 liters 850 = 50 * n = > n = 17 here n = number of cycles . total time = 17 * 3 = 51 as in one cycle there are 3 minutes . thus 51 minutes answer : b" | a = 40 + 30
b = a - 20
c = 850 / b
d = c * 3
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a ) 240 , b ) 75 , c ) 110 , d ) 120 , e ) 200 | d | add(divide(subtract(add(divide(subtract(240, 80), const_2), 80), 80), const_2), 80) | a picnic attracts 240 persons . there are 80 more men than women , and 80 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 80 i . e . y + ( y + 80 ) = 240 i . e . y = 80 i . e . adult = 80 + 80 = 160 adults include only men and women i . e . men + women = 160 let women , w = x then men , m = x + 80 i . e . x + ( x + 20 ) = 2 x + 80 = 160 i . e . x = 40 i . e . men , m = 40 + 80 = 120 answer : option d" | a = 240 - 80
b = a / 2
c = b + 80
d = c - 80
e = d / 2
f = e + 80
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a ) 10 minutes , b ) 1.5 hours , c ) 9 hours and 12 minutes , d ) 15 hours , e ) 1 hour and 48 minutes . | a | inverse(add(divide(const_1, divide(36, 3)), divide(const_1, multiply(20, 3)))) | tap a runs 3 liters of water per minute into a bucket that has a total volume of 36 liters . tap b fills a third of the bucket in 20 mins . working together how long will it take both taps to fill the bucket ? | pipe a has a rate of 3 liters per minute which equals a rate of 180 liters per hour . pipe b has a rate of 36 liters per hour . 180 + 36 ( a + b ) = together they have a rate of 2000 liters per hour . 36 ( the bucket ) / 216 = 1 / 6 = 10 minutes answer - a | a = 36 / 3
b = 1 / a
c = 20 * 3
d = 1 / c
e = b + d
f = 1/(e)
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a ) a ) 70 , b ) b ) 35 , c ) c ) 385 , d ) d ) 280 , e ) e ) 140 | c | add(const_3, const_4) | what is the smallest integer that is multiple of 5 , 7,11 | "it is the lcm of 5 , 7 and 11 which is 385 . the answer is c ." | a = 3 + 4
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a ) rs . 2250 , b ) rs . 3375 , c ) rs . 6000 , d ) rs . 5625 , e ) none of these | c | multiply(22000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 22000 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 Γ ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) Γ 22000 ] = rs . 6000 answer : option c" | a = 2 / 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 1
f = 1/(e)
g = 22000 * f
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a ) 10 % decrease , b ) 20 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 64 % decrease | e | subtract(const_100, multiply(power(divide(40, const_100), const_2), const_100)) | if the radius of a circle is decreased 40 % , what happens to the area ? | "area of square = pi * radius ^ 2 new radius = 0.6 * old radius so new area = ( 0.6 ) ^ 2 old area = > 4 / 25 of old area = > 36 % old area ans : e" | a = 40 / 100
b = a ** 2
c = b * 100
d = 100 - c
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 9 | b | divide(9, divide(multiply(9, subtract(9, const_3)), const_2)) | what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular 9 - sided polygon ) ? | "from any vertex , there are two vertices on sides , which do not make a diagonal but a side . so the remaining n - 3 vertices make diagonals . there are 2 of these diagonals which are the shortest . the probability of choosing one of the shortest diagonals is 2 / 6 = 1 / 3 . the answer is b ." | a = 9 - 3
b = 9 * a
c = b / 2
d = 9 / c
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a ) 14 % , b ) 25 % , c ) 28 % , d ) 32.5 % , e ) 50 % | d | multiply(divide(add(multiply(divide(10, const_100), 30), 10), add(30, 10)), const_100) | if 10 gallons of grape juice are added to 30 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ? | "official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 30 Γ 0.10 = 3 gallons of grape juice . 30 Γ 0.90 = 27 gallons of other components . if we add 10 gallons of grape juice , we will end up with 13 gallons of grape juice and 36 gallons of other components , and we will have a total of 40 gallons of the mixture . so 13 / 40 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 32,5 % . the correct answer is choice ( d )" | a = 10 / 100
b = a * 30
c = b + 10
d = 30 + 10
e = c / d
f = e * 100
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a ) 16 , b ) 20 , c ) 24 , d ) 28 , e ) 32 | c | multiply(12, const_2) | if ( 10 ^ 4 * 3.456789 ) ^ 12 is written as a single term , how many digits would be to the right of the decimal place ? | "3.456789 ^ 12 has 6 * 12 = 72 decimal places . 10 ^ 48 moves the decimal place to the right 48 places . ( 10 ^ 4 * 3.456789 ) ^ 12 has 72 - 48 = 24 digits after the decimal point . the answer is c ." | a = 12 * 2
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a ) 312 cm , b ) 381 cm , c ) 350 cm , d ) 310 cm , e ) 354 cm | b | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 150), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 150 inches into centimeter ? | "1 inch = 2.54 cm 150 inches = 110 * 2.54 = 381 cm answer is b" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 150
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
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a ) 22 , b ) 60 , c ) 88 , d ) 26 , e ) 18 | b | divide(multiply(subtract(const_1, divide(25, const_100)), subtract(50, 30)), divide(25, const_100)) | one day eesha started 30 min late from home and reached her office 50 min late while driving 25 % slower than her usual speed . how much time in min does eesha usually take to reach her office from home ? | we know that speed is inversely proportional to time while she drives 25 % slower means she drove at 3434 ( s ) we know that d = s x t when speed became 3434 ( s ) then time taken should be 4343 ( t ) i . e , she has taken 4343 ( t ) - t extra to cover the distance . extra time = t 3 t 3 = 20 min ( as 20 min late due to slow driving ) actual time t = 60 minutes answer : b | a = 25 / 100
b = 1 - a
c = 50 - 30
d = b * c
e = 25 / 100
f = d / e
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a ) 14 , b ) 16 , c ) 18 , d ) 24 , e ) 34 | b | add(add(power(add(add(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(24, const_10), const_2), const_4), const_2), const_2))) | the sum of three consecutive even numbers is 24 . find the middle number of the three ? | "three consecutive even numbers ( 2 p - 2 ) , 2 p , ( 2 p + 2 ) . ( 2 p - 2 ) + 2 p + ( 2 p + 2 ) = 24 6 p = 24 = > p = 4 . the middle number is : 4 p = 16 . answer : b" | a = 24 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 24 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 24 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 24 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 5.55 sec , b ) 4.55 sec , c ) 5.95 sec , d ) 6.55 sec , e ) 9.55 sec | a | multiply(multiply(multiply(const_0_2778, subtract(50, 40)), 50), inverse(multiply(const_0_2778, add(50, 40)))) | two trains of equal length , running with the speeds of 50 and 40 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 50 - 40 = 10 * 5 / 18 = 50 / 18 t = 50 d = 50 * 50 / 18 = 1250 / 9 rs = 50 + 40 = 90 * 5 / 18 = 25 t = 1250 / 9 * 1 / 25 = 5.55 sec answer : a" | a = 50 - 40
b = const_0_2778 * a
c = b * 50
d = 50 + 40
e = const_0_2778 * d
f = 1/(e)
g = c * f
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a ) a ) 34.7 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.5 , e ) e ) 22 | a | subtract(subtract(40, divide(30, const_100)), divide(30, 6)) | daniel went to a shop and bought things worth rs . 40 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 40 given that out of this rs . 40 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % β΄ ( 40 β 30 / 100 β x ) 6 / 100 = 30 / 100 β 6 ( 40 β 0.3 β x ) = 30 β ( 40 β 0.3 β x ) = 5 β x = 40 β 0.3 β 5 = 34.7 ans - a" | a = 30 / 100
b = 40 - a
c = 30 / 6
d = b - c
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a ) 1 day , b ) 2.1 days , c ) 3 days , d ) 4.5 days , e ) 5 days | d | divide(36, multiply(divide(48, multiply(6, 2)), 2)) | if 6 men can colour 48 m long cloth in 2 days , then 2 men can colour 36 m long cloth in | "the length of cloth painted by one man in one day = 48 / 6 Γ 2 = 4 m no . of days required to paint 36 m cloth by 6 men = 36 / 4 Γ 2 = 4.5 day . d" | a = 6 * 2
b = 48 / a
c = b * 2
d = 36 / c
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a ) 3 , b ) 4.5 , c ) 4 , d ) d ) 10 , e ) e ) 5 | d | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts five hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 5 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 5 + t ) = 40 * t t = 15 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 15 / 60 = 10 hrs time taken by a = 5 + t = 5 + 15 = 20 hrs time taken by k = 10 hrs so k starts 20 - 10 = 10 hrs after a . ( answer d ) | a = 40 - 30
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a ) 10 ^ 8 , b ) 10 ^ 9 , c ) 10 ^ 10 , d ) 10 ^ 11 , e ) 10 ^ 12 | d | divide(0.1, power(0.001, 4)) | the decimal 0.1 is how many times greater than the decimal ( 0.001 ) ^ 4 ? | "0.1 = 10 ^ - 1 ( 0.001 ) ^ 4 = ( 10 ^ - 3 ) ^ 4 = 10 ^ - 12 10 ^ 11 * 10 ^ - 12 = 10 ^ - 1 the answer is d ." | a = 0 ** 1
b = 0 / 1
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a ) 5 seconds , b ) 4.5 seconds , c ) 10 seconds , d ) 2.5 seconds , e ) none of these | c | divide(400, multiply(144, const_0_2778)) | in what time will a train 400 meters long cross an electric pole , if its speed is 144 km / hr | "explanation : first convert speed into m / sec speed = 144 * ( 5 / 18 ) = 40 m / sec time = distance / speed = 400 / 40 = 10 seconds answer : c" | a = 144 * const_0_2778
b = 400 / a
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a ) 106 , b ) 107 , c ) 397 , d ) 109 , e ) 110 | c | add(108, divide(108, 2)) | p software has coding line 5 % more than n , n software has coding line 5 / 2 more than m . m software has 108 lines of coding . find p lines . | "m s / w has 108 line of code n s / w has = 108 + 108 * 5 / 2 = 378 line of code p s / w 5 % more n ' code 378 + 18.9 = 396.9 or 397 line of code answer : c" | a = 108 / 2
b = 108 + a
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a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 11 , d ) 4 / 21 , e ) 5 / 21 | c | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 2), 21), const_4.0), const_2), 2), const_3)) | a certain list consists of 21 different numbers . if n is in the list and n is 2 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction of the sum of the 21 numbers in the list ? | "series : a 1 , a 2 . . . . a 20 , n sum of a 1 + a 2 + . . . + a 20 = 20 * x ( x = average ) so , n = 2 * x hence , a 1 + a 2 + . . + a 20 + n = 22 x so , the fraction asked = 2 x / 22 x = 1 / 11 answer c" | a = 1 * 1
b = 20 / 2
c = b + 21
d = c / 4
e = d * 2
f = e - 2
g = f - 3
h = a / g
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a ) 328897 , b ) 120000 , c ) 877888 , d ) 116000 , e ) 188871 | d | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 20,000 is made ? | "the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 20,000 is made = 96,000 + 20,000 = rs . 1 , 16,000 answer : d" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 20 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 300 , b ) 420 , c ) 560 , d ) 320 , e ) 400 | c | add(multiply(60, 8), multiply(subtract(60, 20), multiply(8, divide(25, const_100)))) | mary works in a restaurant a maximum of 60 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 40 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 40 ) = $ 400 . the total amount is $ 160 + $ 400 = $ 560 answer c 560 . | a = 60 * 8
b = 60 - 20
c = 25 / 100
d = 8 * c
e = b * d
f = a + e
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a ) 30 , b ) 36 , c ) 50 , d ) 44 , e ) 60 | b | divide(600, multiply(subtract(64, 4), const_0_2778)) | how many seconds will a 600 m long train take to cross a man walking with a speed of 4 km / hr in the direction of the moving train if the speed of the train is 64 km / hr ? | "speed of train relative to man = 64 - 4 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 600 * 3 / 50 = 36 sec . answer : b" | a = 64 - 4
b = a * const_0_2778
c = 600 / b
|
a ) 4 : 5 , b ) 9 : 25 , c ) 5 : 4 , d ) 25 : 9 , e ) can not be determined from the information provided | c | divide(circumface(divide(50, const_2)), circumface(divide(40, const_2))) | two interconnected , circular gears travel at the same circumferential rate . if gear a has a diameter of 40 centimeters and gear b has a diameter of 50 centimeters , what is the ratio of the number of revolutions that gear a makes per minute to the number of revolutions that gear b makes per minute ? | "same circumferential rate means that a point on both the gears would take same time to come back to the same position again . hence in other words , time taken by the point to cover the circumference of gear a = time take by point to cover the circumference of gear b time a = 2 * pi * 25 / speed a time b = 2 * pi * 20 / speed b since the times are same , 50 pi / speed a = 40 pi / speed b speeda / speed b = 50 pi / 30 pi = 5 / 4 correct option : c" | a = 50 / 2
b = circumface / (
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a ) 7 , b ) 13 , c ) 15 , d ) 16 , e ) 23 | b | power(divide(10, 5), const_4) | if ( 2 ^ 13 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ? | "given 2 ^ 13 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 13 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans b on comparing the power of 2 = > m = 13" | a = 10 / 5
b = a ** 4
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a ) 15120 , b ) 14500 , c ) 14900 , d ) 14300 , e ) 14000 | a | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 36,000 , what will be the amount a receives ? | total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 β 3 x + 14000 = 50000 β 3 x = 50000 β 14000 = 36000 β x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a , b and c = 21000 : 17000 : 12000 = 21 : 17 : 12 share of a = total profit Γ 21 / 50 = 36000 Γ 21 / 50 = 15,120 answer is a . | a = 5000 * 10
b = 4000 + 5000
c = b + 5000
d = a - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 3 * 12
i = h * 1000
j = 5000 * 10
k = i / j
l = g * k
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a ) 10.00 , b ) 15.00 , c ) 11.50 , d ) 10.50 , e ) 9.75 | a | multiply(500, divide(20, 1000)) | a glucose solution contains 20 grams of glucose per 1000 cubic centimeters of solution . if 500 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | "we are given that a glucose solution contains 20 grams of glucose per 1000 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of glucose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of glucose would be in the container when we have 500 cubic centimeters of solution , we can set up a proportion . we can say : β 20 grams of glucose is to 1000 cubic centimeters of solution as x grams of glucose is to 500 cubic centimeters of solution . β let β s now set up the proportion and solve for x . 20 / 1000 = x / 500 when we cross multiply we obtain : ( 20 ) ( 500 ) = 1000 x 10000 = 1000 x 10 = x there are 10.00 grams of glucose in the solution in the container . the answer is a ." | a = 20 / 1000
b = 500 * a
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a ) 54 , b ) 432 , c ) 2,160 , d ) 2,916 , e ) 148,824 | c | multiply(multiply(6, subtract(6, const_1)), multiply(9, 6)) | right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 6 β€ x β€ 2 and 4 β€ y β€ 9 , then how many different triangles can be drawn that will meet these conditions ? | "we have the rectangle with dimensions 9 * 6 ( 9 horizontal dots and 6 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 6 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot is already occupied by a ) ; choose the y coordinate for vertex b ( as x coordinate is fixed by a ) : 5 c 1 , ( 6 - 1 = 5 as 1 vertical dot is already occupied by a ) . 9 c 1 * 6 c * 8 c 1 * 5 c 1 = 2160 . answer : c ." | a = 6 - 1
b = 6 * a
c = 9 * 6
d = b * c
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a ) 83.08 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | a | subtract(multiply(divide(6000, 756), 10.5), divide(1, 4)) | the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 6000 , brokerage being 1 / 4 % , is : | face value = rs . 6000 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 6000 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs . 100 . = market value of rs . 100 stock + brokerage per rs . 100 = rs . 83.33 . = market value of rs . 100 stock = rs . 83.33 - re . 0.25 . = market value of rs . 100 stock = rs . 83.08 answer : a | a = 6000 / 756
b = a * 10
c = 1 / 4
d = b - c
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a ) 1 cm , b ) 4 cm , c ) 2 cm , d ) 3 cm , e ) 5 cm | b | sqrt(divide(64, const_4)) | the curved surface of a sphere is 64 pi cm 2 . find its radius ? | "4 Γ― β¬ r 2 = 64 = > r = 4 answer b" | a = 64 / 4
b = math.sqrt(a)
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a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 12 , d ) 1 / 4 , e ) 3 / 4 | c | divide(const_1, multiply(4, 3)) | a shopping cart contains only apples , oranges , and pears . if there are 3 times as many oranges as apples , and 4 times as many pears as oranges , then the apples are equivalent to what fraction of the pears ? | o = 3 a p = 4 o = 12 a a = p / 12 the answer is c . | a = 4 * 3
b = 1 / a
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a ) rs . 3360.80 , b ) rs . 3370.80 , c ) rs . 3320.70 , d ) rs . 3230.80 , e ) rs . 3130.80 | b | add(divide(370.8, subtract(power(add(const_1, divide(6, const_100)), const_2), const_1)), 370.8) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 6 % p . a . was rs . 370.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned . | let the sum be rs . p p { [ 1 + 6 / 100 ] 2 - 1 } = 370.80 p ( 6 / 100 ) ( 2 + 6 / 100 ) = 370.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 246 / ( 0.05 ) ( 2.05 ) = 3000 . amount = rs . 3370.80 . answer : b | a = 6 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 370 / 8
f = e + 370
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a ) 0 , b ) 10 , c ) 20 , d ) 30 , e ) 40 | d | multiply(subtract(multiply(subtract(multiply(3, 3), 3), 3), 3), subtract(3, const_1)) | q ' = 3 q - 3 , what is the value of ( 3 ' ) ' ? | "( 3 ' ) ' = ( 3 * 3 - 3 ) ' = 6 ' = 6 * 6 - 6 = 30 answer d" | a = 3 * 3
b = a - 3
c = b * 3
d = c - 3
e = 3 - 1
f = d * e
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['a ) 125 cm 3', 'b ) 625 cm 3', 'c ) 15525 cm 3', 'd ) 1225 cm 3', 'e ) none'] | b | multiply(volume_cube(5), add(const_4, const_1)) | five equal cubes , each of side 5 cm , are placed adjacent to each other . the volume of the new solid formed will be | solution the new solid formed is a cuboid of length 25 cm , breadth 5 cm and height 5 cm β΄ volume = ( 25 x 5 x 5 ) cm 3 βΉ = βΊ 625 cm 3 answer b | a = volume_cube * (
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a ) rs . 1200 , b ) rs . 1100 , c ) rs . 1300 , d ) rs . 1400 , e ) rs . 1000 | a | divide(multiply(180, const_100), subtract(add(const_100, 5), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 180 more , there would have been a gain of 5 % . what is the cost price ? | "90 % 105 % - - - - - - - - 15 % - - - - 180 100 % - - - - ? = > rs . 1200 answer : a" | a = 180 * 100
b = 100 + 5
c = 100 - 10
d = b - c
e = a / d
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a ) 16 kmph , b ) 88 kmph , c ) 54 kmph , d ) 48 kmph , e ) 19 kmph | d | multiply(const_3_6, divide(80, 6)) | a train 80 m in length crosses a telegraph post in 6 seconds . the speed of the train is ? | "s = 80 / 6 * 18 / 5 = 48 kmph answer : d" | a = 80 / 6
b = const_3_6 * a
|
a ) 11 , b ) 77 , c ) 30 , d ) 20 , e ) 34 | d | divide(add(7, 33), const_2) | a man can row upstream at 7 kmph and downstream at 33 kmph , and then find the speed of the man in still water ? | "us = 7 ds = 33 m = ( 33 + 7 ) / 2 = 20 answer : d" | a = 7 + 33
b = a / 2
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | b | subtract(add(55, 85), subtract(150, 30)) | in a class of 150 students 55 speak english ΝΎ 85 speak telugu and 30 speak neither english nor telugu how many speak both english and telugu ? | total = 150 english = 55 telugu = 85 none / neither english nor telugu = 30 no . of students who speak only one or both = 150 - 30 = 120 so , 120 stud ' s can speak english or telugu or both . then , no . of stud ' s speak only english = 120 - 85 ( telugu ) = 35 ` ` ` ` ` ` ` ` ` ` telugu = 120 - 55 = 65 remaining 120 - ( 65 + 35 ) = 20 stud ' s can speak both . b ) | a = 55 + 85
b = 150 - 30
c = a - b
|
a ) 280 , b ) 400 , c ) 540 , d ) 550 , e ) 840 | d | divide(divide(divide(110, subtract(const_1, divide(3, 5))), divide(3, 4)), divide(2, 3)) | of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year . if 110 geese survived the first year and if no more than one goose hatched from each egg , how many goose eggs were laid at the pond ? | "of the goose eggs laid at a certain pond , 2 / 3 hatched and 3 / 4 of the geese that hatched from those eggs survived the first month : 2 / 3 * 3 / 4 = 1 / 2 survived the first month . of the geese that survived the first month , 3 / 5 did not survive the first year : ( 1 - 3 / 5 ) * 1 / 2 = 1 / 5 survived the first year . 110 geese survived the first year : 1 / 5 * ( total ) = 110 - - > ( total ) = 550 . answer : d ." | a = 3 / 5
b = 1 - a
c = 110 / b
d = 3 / 4
e = c / d
f = 2 / 3
g = e / f
|
a ) 250 , b ) 276 , c ) 280 , d ) 285 , e ) 295 | d | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 510 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "explanation : since the month begins with a sunday , to there will be five sundays in the month . required average = ( 510 Γ 5 + 240 Γ 25 / 30 ) = 8550 / 30 = 285 answer : d" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 510
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 29997 , b ) 28088 , c ) 13500 , d ) 19000 , e ) 2799 | c | subtract(15000, 1500) | the price of a t . v . set worth rs . 15000 is to be paid in 20 installments of rs . 1500 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 1500 balance payment = ( 15000 - 1500 ) = rs . 13500 . answer : c" | a = 15000 - 1500
|
a ) 10 , b ) 30 , c ) 15 , d ) 20 , e ) 25 | c | multiply(multiply(multiply(const_2, const_2), const_2), 5) | a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of 5 cubic feet . if a large block has three times the width , twice the depth , and half the length of a normal block , what is the volume of cheese in a large block in cubic feet ? | "volume of cube = lbh = 5 new cube l , b , h are increases of 3 b , 2 h and a decrease of . 5 l new volume of cube = . 5 l * 3 b * 2 h = 3 * lbh = 3 * 5 = 15 answer : c" | a = 2 * 2
b = a * 2
c = b * 5
|
a ) 40 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 % | c | multiply(divide(240, divide(add(120, 240), divide(60, const_100))), const_100) | after a storm deposits 120 billion gallons of water into the city reservoir , the reservoir is 60 % full . if the original contents of the reservoir totaled 240 billion gallons , the reservoir was approximately what percentage full before the storm ? | "when the storm deposited 120 billion gallons , volume of water in the reservoir = 240 + 120 = 360 billion gallons if this is only 60 % of the capacity of the reservoir , the total capacity of the reservoir = 360 / 0.6 = 600 billion gallons therefore percentage of reservoir that was full before the storm = ( 240 / 600 ) * 100 = 40 % option c" | a = 120 + 240
b = 60 / 100
c = a / b
d = 240 / c
e = d * 100
|
a ) 55 , b ) 77 , c ) 56 , d ) 67 , e ) 98 | b | add(65, const_1) | the average of first six prime numbers greater than 65 is ? | "67 + 71 + 73 + 79 + 83 + 89 = 462 / 6 = 77 answer : b" | a = 65 + 1
|
a ) $ 75 , b ) $ 40 , c ) $ 50 , d ) $ 80 , e ) $ 100 | a | subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(25, const_100))) | crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 25 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it before the decrease ? | "deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $ 100 . they had this same selling price , $ 100 , before they made the change , and had a profit of 25 % , so the manufacturing must have been $ 75 . answer = ( a ) ." | a = 50 / 100
b = 50 / a
c = 50 / 100
d = 50 / c
e = 25 / 100
f = d * e
g = b - f
|
a ) 28 , b ) 30 , c ) 22 , d ) 20 , e ) 32 | b | subtract(add(divide(20, const_4), const_4), add(divide(20, const_4), const_4)) | ' a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the addition of the maximum and minimum possible values of ' a + b ' ? | "possible values of a and b can be 5,4 ; 4,5 ( which are same for a + b ) and 1,20 ; 20,1 ( same result for a + b ) so 21 + 9 = 30 . ans b ." | a = 20 / 4
b = a + 4
c = 20 / 4
d = c + 4
e = b - d
|
a ) 12 hours , b ) 24 hours , c ) 36 hours , d ) 10 hours , e ) 15 hours | b | add(multiply(add(add(9, 1.5), subtract(9, 1.5)), 105), multiply(subtract(add(divide(105, add(9, 1.5)), divide(105, subtract(9, 1.5))), add(add(9, 1.5), subtract(9, 1.5))), const_60)) | speed of a boat in standing water is 9 kmph and speed of the stream is 1.5 kmph . a man can 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 upstream = 7.5 kmph speed downstream = 10.5 kmph total time taken = 105 / 7.5 + 105 / 10.5 = 24 hours answer is b" | a = 9 + 1
b = 9 - 1
c = a + b
d = c * 105
e = 9 + 1
f = 105 / e
g = 9 - 1
h = 105 / g
i = f + h
j = 9 + 1
k = 9 - 1
l = j + k
m = i - l
n = m * const_60
o = d + n
|
a ) 6 , b ) 8 , c ) 12 , d ) 4 , e ) 13 | a | divide(30, multiply(add(15, 3), const_0_2778)) | 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 30 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 30 / 5 = 6 seconds . answer : a" | a = 15 + 3
b = a * const_0_2778
c = 30 / b
|
a ) 24,602 , b ) 25,000 , c ) 24,600 , d ) 24,628 , e ) 24,6012 | b | divide(62.50, divide(const_4, 4)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 62.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 62.50 32 x - 31 x = 6250 * 4 x = 25,000 . answer : b" | a = 4 / 4
b = 62 / 50
|
a ) a ) 1055 , b ) b ) 1075 , c ) c ) 1145 , d ) d ) 1070 , e ) e ) 1080 | c | add(multiply(8, 70), multiply(9, 65)) | harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 65 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 Γ 8 = 560 . cost of 9 kg of mangoes = 65 Γ 9 = 585 . total cost he has to pay = 560 + 585 = 1145 . c )" | a = 8 * 70
b = 9 * 65
c = a + b
|
a ) 10 , b ) 30 , c ) 20 , d ) 15 , e ) 40 | b | multiply(20, divide(75, 25)) | it was calculated that 75 men could complete a piece of work in 20 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "before : one day work = 1 / 20 one man β s one day work = 1 / ( 20 * 75 ) now : no . of workers = 50 one day work = 50 * 1 / ( 20 * 75 ) the total no . of days required to complete the work = ( 75 * 20 ) / 50 = 30 answer : b" | a = 75 / 25
b = 20 * a
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | the number of diagonals of a polygon of n sides is given by the formula z = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | "z = n ( n - 3 ) z = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b" | a = 3 + 2
|
a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 8000 , 24000 | e | multiply(subtract(9, const_2), divide(32000, add(3, subtract(9, const_2)))) | divide rs . 32000 in the ratio 3 : 9 ? | "3 / 12 * 32000 = 8000 9 / 12 * 32000 = 24000 answer : e" | a = 9 - 2
b = 9 - 2
c = 3 + b
d = 32000 / c
e = a * d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | power(2, const_2) | what will be the remainder if 2 ^ 300 is divided by 4 ? | d 4 now if you go on calculating 2 ^ 300 , you will lose so much time and it might not even be feasible to carry out so long of calculations . thus we will make use of a trick here . we will calculate the remainder of each power of 2 till we come across a pattern . 2 ^ 1 divided by 4 leaves the remainder 2 . 2 ^ 2 divided by 4 leaves the remainder 0 . 2 ^ 3 divided by 4 leaves the remainder 0 . 2 ^ 4 divided by 4 leaves the remainder 0 . you can see that all the following powers of two will be divisible by 4 . therefore the remainder when 2 ^ 300 is divided by 4 will be 0 only . | a = 2 ** 2
|
a ) 22 , b ) 28 , c ) 77 , d ) 99 , e ) 351 | e | multiply(divide(multiply(13, add(5, const_1)), subtract(multiply(4, 7), multiply(13, 2))), add(7, 2)) | a movie buff owns movies on dvd and on blu - ray in a ratio of 7 : 2 . if she buys 5 more blu - ray movies , that ratio would change to 13 : 4 . if she owns movies on no other medium , what was the original number of movies in her library before the extra purchase ? | if u can just keep an eye on the options 99 is the only multiple of 9 in options given . . so you can mark it wid in seconds . now coming to the process m ( d ) = 7 x and b ( d ) = 2 x now from the next line the new eqn becomes 7 x / ( 2 x + 6 ) = 13 / 4 solving it 28 x = 26 x + 78 x = 39 which means m ( d ) = 273 and b ( d ) = 78 so total initially is m ( d ) + b ( d ) = 351 e | a = 5 + 1
b = 13 * a
c = 4 * 7
d = 13 * 2
e = c - d
f = b / e
g = 7 + 2
h = f * g
|
a ) 22,36 , 63 , b ) 27,45 , 63 , c ) 27,54 , 63 , d ) 54 , 45 , 36 , e ) 27,36 , 63 | e | add(multiply(multiply(1.5, 3.5), const_100), multiply(2, 3.5)) | three numbers are in the ratio 1.5 : 2 : 3.5 and their h . c . f is 18 . the numbers are : | "let the required numbers be 1.5 x , 2 x and 3.5 x . then , their h . c . f = x . so , x = 18 . the numbers are 27,36 , 63 . answer : e" | a = 1 * 5
b = a * 100
c = 2 * 3
d = b + c
|
a ) 5999 , b ) 6788 , c ) 6250 , d ) 4559 , e ) 6989 | c | floor(divide(4500, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100)))) | 10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 4500 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 80 / 100 ) = 4500 x = 6250 answer : c" | a = 100 - 10
b = a / 100
c = 100 - 20
d = c / 100
e = b * d
f = 4500 / e
g = math.floor(f)
|
a ) 12 % , b ) 14 % , c ) 16 % , d ) 20 % , e ) 17.5 % | e | subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 25)), const_100)) | the tax on a commodity is diminished by 25 % but its consumption is increased by 10 % . find the decrease percent in the revenue derived from it ? | "explanation : 100 * 100 = 10000 75 * 110 = 8250 10000 - - - - - - - 1750 100 - - - - - - - ? = 17.5 % e )" | a = 100 + 10
b = 100 - 25
c = a * b
d = c / 100
e = 100 - d
|
a ) 4,700 , b ) 7,500 , c ) 10,500 , d ) 15,000 , e ) 19,600 | a | multiply(divide(add(200, 101), const_2), add(divide(subtract(200, 101), const_2), const_1)) | the sum of the first 60 positive even integers is 3,750 . what is the sum of the odd integers from 101 to 200 , inclusive ? | "101 + 103 + . . . . . . . 199 if we remove 100 from each of these it will be sum of 1 st 100 odd numbers . so 101 + 103 + . . . . . . . 199 = 60 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) sum of 1 st 100 natural numbers = ( 100 * 101 ) / 2 = 5050 sum of 1 st 60 positive even integers = 3750 sum of 1 st 100 odd numbers = 5050 - 3750 = 1300 so 101 + 103 + . . . . . . . 199 = 60 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) = 6000 + 1300 = 4700 a is the answer ." | a = 200 + 101
b = a / 2
c = 200 - 101
d = c / 2
e = d + 1
f = b * e
|
['a ) 54.9 %', 'b ) 58.7 %', 'c ) 62.5 %', 'd ) 66.3 %', 'e ) 70.4 %'] | e | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(6)))) | a wooden cube whose edge length is 6 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ? | there are a total of 6 * 6 * 6 = 216 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 4 * 4 * 4 = 64 cubes . the number of cubes with at least one side painted red is 216 - 64 = 152 cubes the probability that a cube has at least one side painted red is 152 / 216 which is about 70.4 % the answer is e . | a = 1 * 4
b = volume_cube / (
c = 1 - b
d = 100 * c
|
a ) 12,28 , b ) 15,33 , c ) 16,32 , d ) 18,34 , e ) 19,34 | b | subtract(divide(subtract(add(multiply(6, 6), 18), 6), subtract(6, const_1)), const_1) | ages of two persons differ by 18 years . if 6 year ago , the elder one be 3 times as old the younger one , find their present age | "explanation : let the age of younger person is x , then elder person age is ( x + 18 ) = > 3 ( x - 6 ) = ( x + 18 - 6 ) [ 6 years before ] = > 3 x - 18 = x + 12 = > x = 15 . so other person age is x + 18 = 33 answer : option b" | a = 6 * 6
b = a + 18
c = b - 6
d = 6 - 1
e = c / d
f = e - 1
|
a ) 420 , b ) 430 , c ) 312 , d ) 322 , e ) none | a | multiply(28, 15) | the h . c . f . of two numbers is 28 and the other two factors of their l . c . m . are 12 and 15 . the larger of the two numbers is | "solution clearly , the numbers are ( 28 x 12 ) and ( 28 x 15 ) . larger number = ( 28 x 15 ) = 420 . answer a" | a = 28 * 15
|
a ) 12.08 , b ) 12.18 , c ) 12.0 , d ) 6.19 , e ) 6.21 | a | divide(subtract(260, multiply(8, 2.3)), 20) | in the first 8 overs of a cricket game , the run rate was only 2.3 what should be the rate in the remaining 20 overs to reach the target of 260 runs ? | "required run rate = [ 260 - ( 2.3 * 8 ) ] / 20 = 241.60 / 20 = 12.08 answer : a" | a = 8 * 2
b = 260 - a
c = b / 20
|
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