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 ) 6 , b ) 7 , c ) 9.13 , d ) 11 , e ) 13 | c | subtract(divide(1300, 42), divide(120, 6)) | one computer can upload 120 megabytes worth of data in 6 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 42 seconds . how long would it take for the second computer , working on its own , to upload 120 megabytes of data ? | "since the first computer can upload 120 megabytes worth of data in 6 seconds then in 6 * 7 = 42 seconds it can upload 7 * 120 = 840 megabytes worth of data , hence the second compute in 42 seconds uploads 1300 - 840 = 460 megabytes worth of data . the second computer can upload 120 megabytes of data in 9.13 seconds . answer : c ." | a = 1300 / 42
b = 120 / 6
c = a - b
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a ) - 6 , b ) - 3 , c ) 0 , d ) 3 , e ) 6 | c | subtract(add(add(multiply(divide(divide(subtract(306, 18), 6), const_2), 6), 18), 6), add(multiply(divide(divide(subtract(306, 18), 6), const_2), 6), 18)) | set k contains every multiple of 6 from 18 to 306 inclusive . if w is the median of set k , and x is the average ( arithmetic mean ) of set k , what is the value of w - x ? | set k = { 18 , 24,30 , . . . . 306 } let the no . of terms be n . so we know the ap formulae to calculate the the tn term : tn = a + ( n - 1 ) * d where d = common difference of the terms . 306 = 18 + ( n - 1 ) * 6 n = 49 . so the set k consist of n terms . the median of the set having odd nos of elements is ( n + 1 ) / 2 . , which in this case is 25 . let ' s find the 25 th term using the same formulae again : t 25 = 18 + ( 25 - 1 ) * 6 t 25 = 162 so , the median of the set k is 162 i . e . w = 162 . now . lets find the average ( arithmetic mean ) of the set . for that we need to find the sum of all the elements first , lets call it s . since , set k is nothing but a arithmetic progression series having first element ( a ) as 18 , common difference ( d ) as 6 and no . of terms ( n ) as 49 . using the formulae to calculate sum of an ap series , which is s = n / 2 [ 2 a + ( n - 1 ) * d ] , we will calculate the sum . so , s = 49 / 2 [ 2 * 18 + ( 49 - 1 ) * 6 ] this gives us s = 7938 . now arithmetic mean of set k = 7938 / no . of terms = 7938 / 49 = 162 . so x = 162 . now , ( w - x ) = ( 162 - 162 ) = 0 . therefore correct answer = c | a = 306 - 18
b = a / 6
c = b / 2
d = c * 6
e = d + 18
f = e + 6
g = 306 - 18
h = g / 6
i = h / 2
j = i * 6
k = j + 18
l = f - k
|
a ) β 108 , b ) β 44 , c ) 10 , d ) - 17 , e ) 18 | d | divide(add(add(add(multiply(multiply(20, 2), 2), multiply(20, 2)), 20), 4), multiply(4, 2)) | # p is defined as 2 p + 20 for any number p . what is p , if # ( # ( # p ) ) = 4 ? | # p = 2 p + 20 - - - > # ( # p ) = 2 ( 2 p + 20 ) + 20 = 4 p + 60 and thus # ( 4 p + 60 ) = 2 ( 4 p + 60 ) + 20 = 8 p + 140 = 4 - - - > 8 p = - 136 - - - > p = - 17 , d is the correct answer . | a = 20 * 2
b = a * 2
c = 20 * 2
d = b + c
e = d + 20
f = e + 4
g = 4 * 2
h = f / g
|
a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 5 / 3 , e ) 40 / 3 | e | divide(log(9), log(power(3, 0.15))) | if n = 3 ^ 0.15 and n ^ b = 9 , b must equal | "15 / 100 = 3 / 20 n = 3 ^ 3 / 20 n ^ b = 3 ^ 2 ( 3 ^ 3 / 20 ) ^ b = 3 ^ 2 b = 40 / 3 answer : e" | a = math.log(9)
b = 3 ** 0
c = math.log(b)
d = a / c
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a ) 4 : 5 , b ) 4 : 3 , c ) 4 : 6 , d ) 4 : 9 , e ) 4 : 2 | c | divide(sqrt(16), sqrt(36)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 36 hours and 16 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 16 : β 36 = 4 : 6 answer : c" | a = math.sqrt(16)
b = math.sqrt(36)
c = a / b
|
a ) 16 % , b ) 9.66 % , c ) 9.09 % , d ) 9.98 % , e ) 17 % | c | multiply(divide(subtract(60, 55), 55), const_100) | john makes $ 55 a week from his job . he earns a raise andnow makes $ 60 a week . what is the % increase ? | "increase = ( 5 / 55 ) * 100 = ( 1 / 11 ) * 100 = 9.09 % . c" | a = 60 - 55
b = a / 55
c = b * 100
|
a ) 190 , b ) 189 , c ) 180 , d ) 199 , e ) 1981 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | what is the sum of all the prime numbers greater than 10 but less than 40 ? | "required sum = ( 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 ) = 180 note : 1 is not a prime number answer c" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 10
t = 2 + 4
u = s + t
|
a ) rs . 12008 , b ) rs . 12000 , c ) rs . 12002 , d ) rs . 1229 , e ) rs . 12021 | b | divide(8748, subtract(const_1, multiply(divide(10, const_100), 3))) | the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs . 8748 , its purchase price was | explanation : = rs . 12000 answer : b | a = 10 / 100
b = a * 3
c = 1 - b
d = 8748 / c
|
a ) 92 / 198 , b ) 92 / 12 , c ) 92 / 13 , d ) 92 / 10 , e ) 92 / 15 | e | divide(95, 97) | find 95 Γ Γ 97 | "here both numbers are less than 100 . so they are deficient of - 5 and - 3 compared with 100 . so answer : e" | a = 95 / 97
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | divide(add(multiply(factorial(17), factorial(const_4.0)), multiply(factorial(17), factorial(82))), 17) | what is the units digit of 17 ^ 73 Γ 13 ^ 82 Γ 11 ^ 87 ? | "to find : the units digit of 17 ^ 83 Γ 13 ^ 82 Γ 11 ^ 87 let ' s reduce the clutter and simplify the product ( 7 ^ 83 ) ( 3 ^ 82 ) ( 1 ^ 87 ) 7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4 th power so 7 ^ 5 has the same last digit as 7 ^ 1 , 7 ^ 9 , 7 ^ 13 thus , 7 ^ 83 has the same last digit as 7 ^ 3 , 7 ^ 7 , 7 ^ 11 i . e . 3 3 has a cyclicity of 4 : exactly the same routine as above thus , 3 ^ 82 has the same last digit as 3 ^ 2 , 3 ^ 6 , 3 ^ 10 i . e . 9 any power of 1 will result in 1 as the last digit so , product of our last digits = 3 x 9 x 1 = 27 . . . . last digit is 8 correct option : e" | a = math.factorial(17)
b = math.factorial(4)
c = a * b
d = math.factorial(17)
e = math.factorial(82)
f = d * e
g = c + f
h = g / 17
|
a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | a | add(2001, divide(add(divide(70, const_100), subtract(4.40, 4.20)), subtract(divide(30, const_100), subtract(4.40, 4.20)))) | the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 70 cents more than the commodity y ? | "the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 70 = 90 cents on commodity y , which will take 9 years . the answer is a ." | a = 70 / 100
b = 4 - 40
c = a + b
d = 30 / 100
e = 4 - 40
f = d - e
g = c / f
h = 2001 + g
|
a ) 1600 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3400 | a | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(100, 11.25), 6)) | how many bricks , each measuring 100 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 100 x 11.25 x 6 ) = 1600 answer : a" | a = 8 * 100
b = 6 * 100
c = a * b
d = c * 22
e = 100 * 11
f = e * 6
g = d / f
|
a ) 72 / 80 , b ) 72 / 100 , c ) 57 / 80 , d ) 19 / 24 , e ) 76 / 100 | d | multiply(divide(divide(multiply(divide(add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(subtract(const_100, 10), const_100))), const_2), subtract(const_100, 25)), const_100), multiply(subtract(const_100, 20), divide(subtract(const_100, 10), const_100))), const_10) | real - estate salesman z is selling a house at a 20 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 10 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 25 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ? | "let retail price = $ 100 . z sells with 20 % discount . so z sells at $ 80 . x matches the price and gives additional 10 % . so , x sells at 80 - ( 10 % of 80 ) = 80 - 8 = $ 72 y avgs x and z [ ( 72 + 80 ) / 2 = 76 ] . y gives additional 25 % discount . so , y sells at 76 - ( 25 % of 76 ) = 76 - 19 = $ 57 now , ratio of y to x = 57 / 72 = 19 / 24 answer will be d ." | a = 100 - 20
b = 100 - 20
c = 100 - 10
d = c / 100
e = b * d
f = a + e
g = f / 2
h = 100 - 25
i = g * h
j = i / 100
k = 100 - 20
l = 100 - 10
m = l / 100
n = k * m
o = j / n
p = o * 10
|
a ) 19 , b ) 15 , c ) 10 , d ) 8 , e ) 13 | d | multiply(multiply(add(3, divide(subtract(sqrt(28), 3), 2)), divide(subtract(sqrt(28), 3), 2)), 2) | if a - b = 3 and a ^ 2 + b ^ 2 = 28 , find the value of ab . | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 25 - 9 = 16 ab = 8 . answer is d ." | a = math.sqrt(28)
b = a - 3
c = b / 2
d = 3 + c
e = math.sqrt(28)
f = e - 3
g = f / 2
h = d * g
i = h * 2
|
a ) 20 , b ) 50 , c ) 40 , d ) 100 , e ) 25 | a | divide(multiply(10, 40), subtract(40, 20)) | a group of men decided to do a work in 20 days , but 10 of them became absent . if the rest of the group did the work in 40 days , find the original number of men ? | "original number of men = 10 * 40 / ( 40 - 20 ) = 20 answer is a" | a = 10 * 40
b = 40 - 20
c = a / b
|
a ) s . 5266 , b ) s . 5396 , c ) s . 5228 , d ) s . 5218 , e ) s . 52192 | b | divide(8310, add(const_1, divide(multiply(6, 9), const_100))) | mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8310 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8310 = p + ( 54 p / 100 ) 8310 = ( 100 p + 54 p ) / 100 8310 = 154 p / 100 p = ( 8310 * 100 ) / 154 = rs . 5396.104 answer : b" | a = 6 * 9
b = a / 100
c = 1 + b
d = 8310 / c
|
a ) 16 % , b ) 19 % , c ) 17 % , d ) 13 % , e ) 12 % | b | subtract(const_100, divide(multiply(788, const_100), 980)) | an article is bought for rs . 980 and sold for rs . 788 , find the loss percent ? | "980 - - - - 192 100 - - - - ? = > 19 % answer : b" | a = 788 * 100
b = a / 980
c = 100 - b
|
a ) 12 / 13 , b ) 11 / 13 , c ) 10 / 15 , d ) 1 / 15 , e ) 4 / 14 | b | divide(subtract(divide(divide(factorial(15), factorial(subtract(15, const_4))), factorial(const_4)), divide(divide(factorial(subtract(15, const_5)), factorial(subtract(subtract(15, const_5), const_4))), factorial(const_4))), divide(divide(factorial(15), factorial(subtract(15, const_4))), factorial(const_4))) | out of 15 students in a class , 7 are from maharashtra , 5 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ? | total possible ways of selecting 4 students out of 15 = 15 c 4 = ( 15 * 14 * 13 * 12 ) / ( 4 * 3 * 2 * 1 ) = 1365 the number of ways of selecting a 4 students in which no student belongs to karnataka = 10 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 10 c 4 = 1155 probability = 1155 / 1365 = 11 / 13 answer is b | a = math.factorial(15)
b = 15 - 4
c = math.factorial(b)
d = a / c
e = math.factorial(4)
f = d / e
g = 15 - 5
h = math.factorial(g)
i = 15 - 5
j = i - 4
k = math.factorial(j)
l = h / k
m = math.factorial(4)
n = l / m
o = f - n
p = math.factorial(15)
q = 15 - 4
r = math.factorial(q)
s = p / r
t = math.factorial(4)
u = s / t
v = o / u
|
a ) 7 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(const_10, 3) | suppose that f ( x ) is a function such that for every real number x , i ) f ( x ) + f ( 1 - x ) = 10 and ( ii ) f ( 1 + x ) = 3 + f ( x ) . then f ( x ) + f ( - x ) must equal | since ( ii ) holds for every real number x , it will remain valid if we replace x with - x . therefore , f ( 1 - x ) = 3 + f ( - x ) . now , from ( i ) , 10 = f ( x ) + f ( 1 - x ) = f ( x ) + 3 + f ( - x ) so that f ( x ) + f ( - x ) = 10 - 3 = 7 . ( observe that f ( x ) = 3 x + 4 satisfies the conditions in the problem . ) correct answer a | a = 10 - 3
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a ) 220 , b ) 65 , c ) 216 , d ) 219 , e ) 230 | c | multiply(18, 12) | each child has 12 crayons and 24 apples . if there are 18 children , how many crayons are there in total ? | 12 * 18 = 216 . answer is c . | a = 18 * 12
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a ) 8 , b ) 64 , c ) 168 , d ) 102 , e ) 144 | c | subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 90), const_100), 48) | in a certain warehouse , 90 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ? | "if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 90 % of the packages weigh less than 75 pounds so , 90 % oft = number of packages that weigh less than 75 pounds 90 % of 240 = 216 , so 216 packages weigh less than 75 pounds of those 216 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 216 - 48 = 168 = c" | a = 100 - 80
b = 48 / a
c = b * 100
d = c * 90
e = d / 100
f = e - 48
|
a ) 0 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | what is the units digit of the expression 14 ^ 7 β 19 ^ 4 ? | i think answer on this one should be e too . since we know that 14 ^ 7 > 19 ^ 4 , as will said one should always check if the number is positive . | a = 4 + 3
b = a + 2
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['a ) Ο β \u200b 12', 'b ) 9 / Ο', 'c ) 9 β 3 / Ο', 'd ) 6 β 3 / Ο', 'e ) none of these'] | c | multiply(const_3, multiply(sqrt(divide(const_3, const_pi)), sqrt(subtract(divide(12, const_pi), divide(const_3, const_pi))))) | there are two concentric circles such that the area of the outer circle is four times the area of the inner circle . let a , b and c be three distinct points on the perimeter of the outer circle such that ab and ac are tangents to the inner circle . if the area of the outer circle is 12 square centimetres then the area ( in square centimetres ) of the triangle abc would be | explanation : outer circle : since Ο r Β² = 12 , r = β ( 12 / Ο ) = 2 β ( 3 / Ο ) . inner circle : since the area is 1 / 4 the area of the outer circle , Ο r Β² = 3 and r = β ( 3 / Ο ) ao is the radius of the outer circle and od is the radius of the inner circle . a line tangent to a circle and a radius drawn to the point of tangency form a right angle . thus , β aod is a right triangle . hypotenuse ao is twice the length of leg od . a right triangle whose hypotenuse is twice the length of a leg is a 30 - 60 - 90 triangle . thus , β aod is a 30 - 60 - 90 triangle . in a 30 - 60 - 90 triangle , the sides are proportioned x : x β 3 : 2 x . thus , ad = β 3 * β ( 3 / Ο ) * = 3 / β Ο . using the same reasoning for β ocd , we can deduce that cd = 3 / β Ο . thus , ac = 2 * 3 / β Ο = 6 / β Ο . the formula for the area of an equilateral triangle = ( s Β² β 3 ) / 4 . since ac = 6 / β Ο , the area of β abc = ( 6 / β Ο ) Β² * β 3 / 4 = 36 / Ο * β 3 / 4 = ( 9 β 3 ) / Ο . answer : c | a = 3 / math.pi
b = math.sqrt(a)
c = 12 / math.pi
d = 3 / math.pi
e = c - d
f = math.sqrt(e)
g = b * f
h = 3 * g
|
a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | add(const_2, const_3) | if 425 / 999 = 0.125 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ? | 0 . [ u ] 125 [ / u = 0.425425425 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 5 . since 81 is multiple of 3 , the 81 st digit is 5 . answer e | a = 2 + 3
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a ) a ) 590280 , b ) b ) 590100 , c ) c ) 591900 , d ) d ) 590000 , e ) e ) 927000 | d | add(multiply(multiply(divide(divide(500000, const_2), 10), const_3), const_4), 15000) | a and b start a business , with a investing the total capital of rs . 500000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 15000 per month from the total profit and any profit remaining is equally shared by both of them . at the end of the year , it was found that the income of a is twice that of b . find the total profit for the year ? | "interest received by a from b = 10 % of half of rs . 500000 = 10 % * 250000 = 25000 . amount received by a per annum for being a working partner = 15000 * 12 = rs . 180000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 25000 + 180000 + p ) total income of b = only his share from the remaining profit = ' p ' , as a and b share the remaining profit equally . income of a = twice the income of b ( 25000 + 180000 + p ) = 2 ( p ) p = 205000 total profit = 2 p + 180000 = 2 * 205000 + 180000 = 590000 answer : d" | a = 500000 / 2
b = a / 10
c = b * 3
d = c * 4
e = d + 15000
|
a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 18 | e | multiply(add(2, const_1), add(5, const_1)) | if p and q are prime numbers , how many divisors does the product p ^ 2 * q ^ 5 have ? | "when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is e . 3 * 6 = 18" | a = 2 + 1
b = 5 + 1
c = a * b
|
a ) 5 Β½ ft , b ) 7 Β½ ft , c ) 8 Β½ ft , d ) 9 Β½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least Β½ foot when they are inside the traffic lane , what should be the limit l on the height of vehicles that are allowed to use the tunnel ? | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer l is 8 - . 5 = 7.5 . . . choose ( b ) !" | a = 12 / 2
b = 1 / 2
c = a - b
|
a ) 12.5 % , b ) 13.5 % , c ) 14 % , d ) 15 % , e ) 16 % | a | divide(divide(subtract(subtract(subtract(70000, 42000), 12000), multiply(divide(const_1, const_2), subtract(70000, 42000))), add(15, const_1)), const_10) | a factory has 15 machines of equal efficiency in its factory . the annual manufacturing costs are rs . 42000 & establishment charges are rs . 12000 . the annual output of the company is rs . 70000 . the annual output and manufacturing costs are directly proportional to the number of machines . the shareholders get 12 . 512.5 profit , which is directly proportional to the annual output of the company . if 7 . 147.14 machines remain closed throughout the year , then the % decrease in the amountof profit of the shareholders would be ? | option ( b ) is correct original profit = 70,000 β 42,000 β 12,000 = 16 , 00070,000 β 42,000 β 12,000 = 16,000 if 7.14 % 7.14 % of 14 i . e . one of the machines closed throughout the year , then change in profit will be : = 1314 Γ [ 70,000 β 42,000 ] = 1314 Γ [ 70,000 β 42,000 ] = 14,000 = 14,000 thus , decrease in the profit % % = 200016000 Γ 100 = 200016000 Γ 100 = 12.5 % a | a = 70000 - 42000
b = a - 12000
c = 1 / 2
d = 70000 - 42000
e = c * d
f = b - e
g = 15 + 1
h = f / g
i = h / 10
|
a ) 338 , b ) 276 , c ) 322 , d ) 231 , e ) 121 | b | multiply(23, 12) | the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 11 and 12 . the larger of the two numbers is : | "clearly , the numbers are ( 23 * 11 ) and ( 23 * 12 ) . larger number = ( 23 * 12 ) = 276 . answer : b" | a = 23 * 12
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a ) 2500 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2050 | a | add(3600, divide(multiply(3600, 20), const_100)) | the present population of a town is 3600 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 3600 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3600 / ( 1 + 20 / 100 ) ^ 2 = 3600 / ( 6 / 5 ) ^ 2 = 2500 ( approximately ) answer is a" | a = 3600 * 20
b = a / 100
c = 3600 + b
|
a ) a ) 12 , b ) b ) 24 , c ) c ) 36 , d ) d ) 48 , e ) e ) 60 | b | subtract(20, multiply(multiply(12, const_2.0), 2)) | evaluate : 20 - 12 Γ· 4 Γ 2 = | "according to order of operations , 12 Γ· 4 Γ 2 ( division and multiplication ) is done first from left to right 12 Γ· 4 Γ 2 = 3 Γ 2 = 6 hence 20 - 12 Γ· 4 Γ 2 = 20 - 6 = 14 correct answer is b ) 22" | a = 12 * 2
b = a * 2
c = 20 - b
|
a ) 172 , b ) 173 , c ) 174 , d ) 175 , e ) none of the above | c | add(172, divide(subtract(subtract(multiply(40, 50), multiply(subtract(40, 2), subtract(50, 2))), 172), 2)) | the average of a batsman for 40 innings is 50 runs . his highest score exceeds his lowest score by 172 runs . if these two innings are excluded , his average drops by 2 runs . find his highest score . | total runs = 40 Γ 50 = 2000 let his highest score be = x then his lowest score = x β 172 now 200 β x β ( x β 172 ) / 38 = 48 β 2 x = 2172 β 1824 β x = 174 answer c | a = 40 * 50
b = 40 - 2
c = 50 - 2
d = b * c
e = a - d
f = e - 172
g = f / 2
h = 172 + g
|
a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 14 % | e | multiply(divide(subtract(divide(add(30, 9), 150), divide(9, 50)), divide(9, 50)), const_100) | a corporation paid $ 9 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 million in profits ? | "difference in ratios = ( 30 / 150 ) - ( 9 / 50 ) = ( 1 / 50 ) % change = ( change ( 1 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 14 % answer - e" | a = 30 + 9
b = a / 150
c = 9 / 50
d = b - c
e = 9 / 50
f = d / e
g = f * 100
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a ) 4 m , b ) 5 m , c ) 6 m , d ) 7 m , e ) 8 m | e | power(divide(129024, multiply(multiply(6, 7), 6)), divide(const_1, const_3)) | the height of the wall is 6 times its width and length of the wall is 7 times its height . if the volume of the wall be 129024 cu . m . its width is | "explanation : let width = x then , height = 6 x and length = 42 x 42 x Γ£ β 6 x Γ£ β x = 129024 x = 8 answer : e" | a = 6 * 7
b = a * 6
c = 129024 / b
d = 1 / 3
e = c ** d
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a ) 1000 , b ) 3000 , c ) 7000 , d ) 5000 , e ) 2000 | d | divide(136.09, divide(43.9, 136.09)) | evaluate : 136.09 + 43.9 | "25 / 0.0005 = ( 25 * 10000 ) / ( 0.0005 * 10000 ) = 25000 / 5 = 5000 answer is d" | a = 43 / 9
b = 136 / 9
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a ) 4 / 11 , b ) 5 / 17 , c ) 6 / 25 , d ) 7 / 30 , e ) 8 / 37 | d | multiply(divide(add(const_4, const_4), 16), divide(subtract(add(const_4, const_4), 1), subtract(16, 1))) | 16 balls are numbered 1 to 16 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 8 / 16 p ( 2 nd ball is also even ) = 7 / 15 p ( both balls are even ) = 8 / 16 * 7 / 15 = 7 / 30 the answer is d ." | a = 4 + 4
b = a / 16
c = 4 + 4
d = c - 1
e = 16 - 1
f = d / e
g = b * f
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | subtract(17, 3) | the number of years of service of the 8 employees in a production department are 15 , 10 , 9 , 17 , 6 , 3 , 14 and 16 . what is the range in the number of years of service of the 8 employees ? | = 17 - 3 = 14 answer e | a = 17 - 3
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a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(const_60.0, multiply(const_10, const_2)) | how many factors of 440 are odd numbers greater than 1 ? | when factorized , 440 has 4 prime factors . of these prime factors 2 are odd and 2 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 2 . the total number of odd factors greater than 1 are 2 . ( option a ) | a = 10 * 2
b = const_60 / 0
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a ) 3.625 , b ) 3.5 , c ) 3 , d ) 4 , e ) 4.5 | c | add(divide(subtract(divide(35, const_10), 2.5), const_2), 2.5) | annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.5 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 35 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exactly 35 minutes , for how many kilometers total did she hike east ? | set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.5 from which t = 25 mins . we know total journey time now is 35 + 25 = 60 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 60 = 6 km now the total journey would be halved as distance would be same in each direction . 6 / 2 = 3 c | a = 35 / 10
b = a - 2
c = b / 2
d = c + 2
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a ) 62 degrees , b ) 32 degrees , c ) 65 degrees , d ) 66 degrees , e ) 67 degrees | b | multiply(8, subtract(multiply(4, 59), multiply(4, 58))) | the average temperature of the town in the first 4 days of a month was 58 degrees . the average for the second , third , fourth and fifth days was 59 degrees . if the temperatures of the first and fifth days were in the ratio 7 : 8 , then what is the temperature on the fifth day ? | explanation : sum of temperatures on 1 st , 2 nd , 3 rd and 4 th days = ( 58 * 4 ) = 232 degrees . . . ( 1 ) sum of temperatures on 2 nd , 3 rd , 4 th and 5 th days - ( 59 * 4 ) = 236 degrees . . . . ( 2 ) subtracting ( 1 ) from ( 2 ) , we get : temp , on 5 th day - temp on 1 st day = 4 degrees . let the temperatures on 1 st and 5 th days be 7 x and 8 x degrees respectively . then , 8 x - 7 x = 4 or x = 4 . answer : b temperature on the 5 th day = 8 x = 32 degrees . | a = 4 * 59
b = 4 * 58
c = a - b
d = 8 * c
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a ) 4 , b ) 7 , c ) 8 , d ) 10 , e ) none of these | d | subtract(divide(49, 49), const_1) | 49 Γ£ β 49 Γ£ β 49 Γ£ β 49 x 49 = 7 ? | "49 Γ£ β 49 Γ£ β 49 Γ£ β 49 x 49 = 7 ? or , 7 ( 2 ) Γ£ β 7 ( 2 ) Γ£ β 7 ( 2 ) Γ£ β 7 ( 2 ) x 7 ( 2 ) = 7 ? or 7 ( 10 ) = 7 ? or , ? = 10 answer d" | a = 49 / 49
b = a - 1
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a ) 3.6 , b ) 6 , c ) 18 , d ) 3 , e ) none of these | d | sqrt(divide(multiply(108, const_100), 1200)) | reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 108 as interest at the end of the loan period , what was the rate of interest ? | let rate = r % and time = r years . then , 1200 x r x r / 100 = 108 12 r 2 = 108 r 2 = 9 r = 3 . answer : d | a = 108 * 100
b = a / 1200
c = math.sqrt(b)
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a ) $ 5200 , b ) $ 5408 , c ) $ 5208 , d ) $ 5608 , e ) $ 5808 | b | add(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 4), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 4), const_100)), 4), const_100)) | today joelle opened an interest - bearing savings account and deposited $ 5,000 . if the annual interest rate is 4 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | "interest for 1 st year = 5000 * 4 / 100 = 200 interest for 2 nd year = 5200 * 4 / 100 = 208 total = 5000 + 200 + 208 = 5408 answer : b" | a = 3 * 2
b = a * 100
c = b * 10
d = 3 * 2
e = d * 100
f = e * 10
g = f * 4
h = g / 100
i = c + h
j = 3 * 2
k = j * 100
l = k * 10
m = 3 * 2
n = m * 100
o = n * 10
p = o * 4
q = p / 100
r = l + q
s = r * 4
t = s / 100
u = i + t
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a ) 1 , b ) 1.25 , c ) 1.33 , d ) 1.5 , e ) 1.75 | c | divide(20, add(divide(20, 2), divide(20, 4))) | ana climbs up stairs 20 floors building and back . she takes the exact same route both ways . on the trip up she climbs an average speed of 2 steps per second . on the trip back she climbs down at an average speed of 4 steps per second . what is her approximate average speed for the round trip in miles per hour ? | average speed = total distance / total time here number of steps to 20 th floor and back is same as she takes the same route . d = 2 t 1 and d = 4 t 2 therefore , t 1 = d / 2 and t 2 = d / 4 t 1 + t 2 = 3 d / 4 therefore , average speed = 2 d / 3 d / 4 = 2 d * 4 / 3 d answer is 1.3 since , the options are far away we need not solve till the decimal points so correct answer is option c | a = 20 / 2
b = 20 / 4
c = a + b
d = 20 / c
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a ) 1 : 9 , b ) 1 : 225 , c ) 1 : 52 , d ) 1 : 522 , e ) none | a | divide(const_4, const_100) | a cube of edge 6 cm is cut into cubes each of edge 2 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | "sol . required ratio = 6 * 2 * 2 / 6 * 6 * 6 = 1 / 9 = 1 : 9 . answer a" | a = 4 / 100
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a ) a ) 3 , b ) b ) 1 , c ) c ) 5 , d ) 4 , e ) e ) 28 | d | add(6, 6) | two positive integers differ by 4 , and sum of their reciprocals is 6 . then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 6 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 6 or ( n ^ 2 + 4 n ) * 6 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 4 . hence , d" | a = 6 + 6
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a ) 3 β 7 , b ) 4 β 7 , c ) 3 β 4 , d ) 4 β 3 , e ) 7 β 4 | c | divide(divide(3, 7), divide(4, 7)) | if a mixture is 3 β 7 alcohol by volume and 4 β 7 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 3 / 7 } / { 4 / 7 } = 3 / 4 answer : c" | a = 3 / 7
b = 4 / 7
c = a / b
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a ) 5.2 , b ) 6.9 , c ) 9.6 , d ) 7.8 , e ) 4.7 | c | divide(add(4, 8), add(3, 8)) | the ratio of two quantities is 4 : 3 . if each of the quantities is decreased by 8 , their ratio changes to 3 : 1 then the smallest number is ? | "let the numbers be 4 x and 3 x then 4 x - 8 / 3 x - 8 = 3 / 1 4 x - 8 = 9 x - 24 5 x = 16 x = 3.2 smallest number = 3 * 3.2 = 9.6 answer is c" | a = 4 + 8
b = 3 + 8
c = a / b
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a ) 4 / 19 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 2 1 / 9 | a | subtract(add(divide(4, 19), divide(2, 13)), divide(4, 26)) | the instructions state that cheryl needs 4 / 19 square yards of one type of material and 2 / 13 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 4 / 26 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 4 / 19 + 2 / 13 left part 4 / 26 - - - > 2 / 13 so used part 4 / 19 + 2 / 13 - 2 / 13 = 4 / 19 answer : a" | a = 4 / 19
b = 2 / 13
c = a + b
d = 4 / 26
e = c - d
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a ) 2880 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none of these | a | multiply(multiply(divide(270, 6), 4), 16) | running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 16 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) Γ’ Λ Β΄ 6 Γ£ β 1 Γ£ β x = 16 Γ£ β 4 Γ£ β 270 Γ’ β‘ β x = 16 x 4 x 270 / 6 = 2880 . answer a" | a = 270 / 6
b = a * 4
c = b * 16
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a ) 1232 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(154, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 154 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ? | "a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 154 liters was transferred from container c to container b , then both the containers would have equal quantities of milk . 5 k - 154 = 3 k + 154 = > 2 k = 308 = > k = 154 the initial quantity of milk in a = 8 k = 8 * 154 = 1232 liters ." | a = 1/(2)
b = 62 / 5
c = 1 - b
d = a - c
e = 154 / d
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a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | e | subtract(subtract(170, 60), 30) | of the 170 people at a party , 70 were women , and 30 women tried the appetizer . if 60 people did not try the appetizer , what is the total number of men who tried the appetizer ? | total people at party = 170 women = 70 so men 170 - 70 = 100 no . of pple who tried appetizer = 170 - 60 ( given info ) = 110 no of women who tried appetizer = 30 so remaining ppl ( men ) who tried the appetizer = 110 - 30 = 80 correct option e | a = 170 - 60
b = a - 30
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a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,170 , e ) $ 1,210 | d | subtract(multiply(130, divide(const_100, 10)), 130) | if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 130 , how much more remains to be paid ? | 10 / 100 p = 130 > > p = 130 * 100 / 10 = 1300 1300 - 130 = 1170 answer : d | a = 100 / 10
b = 130 * a
c = b - 130
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a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 4.8 , e ) 2 | a | divide(add(divide(18, 6), divide(36, 6)), const_2) | a man swims downstream 36 km and upstream 18 km taking 6 hours each time , what is the speed of the man in still water ? | "36 - - - 6 ds = 6 ? - - - - 1 18 - - - - 6 us = 3 ? - - - - 1 m = ? m = ( 6 + 3 ) / 2 = 4.5 answer : a" | a = 18 / 6
b = 36 / 6
c = a + b
d = c / 2
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a ) 6 , b ) 10 , c ) 15 , d ) 40 , e ) 22 | c | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(60, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(60, const_10))), const_10)) | a rectangular block 12 cm by 36 cm by 60 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 12 * 36 * 60 = 25920 cm ^ 3 side of the largest cube = h . c . f of 12 , 36,60 = 12 cm volume of the cube = 12 * 12 * 12 = 1728 cm ^ 3 number of cubes = 25920 / 1728 = 15 answer is c" | a = rectangle_area / (
b = a - 10
c = 3 * 2
d = 1000 * c
e = b + d
f = 3 * 1000
g = 60 * 10
h = f + g
i = e / h
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a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | divide(add(divide(35, 5), divide(65, 5)), const_2) | a man swims downstream 65 km and upstream 35 km taking 5 hours each time ; what is the speed of the current ? | "65 - - - 5 ds = 13 ? - - - - 1 35 - - - - 5 us = 7 ? - - - - 1 s = ? s = ( 13 - 7 ) / 2 = 3 answer : b" | a = 35 / 5
b = 65 / 5
c = a + b
d = c / 2
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a ) 621 , b ) 1400 , c ) 236 , d ) 600 , e ) 211 | b | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 3500) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 3500 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 3500 = > 5 parts out of 15 parts is rs . 3500 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 3500 ) = rs . 1400 answer : b" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 3500
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a ) 80 , b ) 40 , c ) 60 , d ) none , e ) can not be determined | a | divide(60, 75) | how many pieces of 75 cm can be cut from a rope 60 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 60 meters long is = ( 60 meters ) / ( 75 cm ) = ( 60 meters ) / ( 0.75 meters ) = 80 answer : a" | a = 60 / 75
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a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 38 % | e | subtract(multiply(divide(subtract(const_100, 8), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 8 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.92 * 150 x = 138 x hence total profit = 38 % option e" | a = 100 - 8
b = a / 100
c = 100 + 20
d = 100 + 25
e = d / 100
f = c * e
g = b * f
h = g - 100
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a ) 27488 , b ) 27424 , c ) 27474 , d ) 27784 , e ) 27468 | c | subtract(31111, subtract(add(3699, 1985), 2047)) | ( ? ) + 3699 + 1985 - 2047 = 31111 | c 27474 x + 3699 + 1985 - 2047 = 31111 x + 3699 + 1985 = 31111 + 2047 x + 5684 = 33158 x = 33158 - 5684 = 27474 . | a = 3699 + 1985
b = a - 2047
c = 31111 - b
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a ) 30 , b ) 22 , c ) 20 , d ) 25 , e ) 100 | d | divide(multiply(10, 5), 2) | pens , pencils and markers in a jenna ' s desk are in the ratio of 2 : 2 : 5 . if there are 10 pens , the number of markers in jenna ' s desk is : | explanation : let pens = 2 x , pencils = 2 x & markers = 5 x . now , 2 x = 10 hence x = 5 . number of markers = 5 x which is 25 . answer : d | a = 10 * 5
b = a / 2
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a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 4000 , e ) $ 12000 | d | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(5, const_100)), const_4)), multiply(const_100, const_100)) | carl is facing very difficult financial times and can only pay the interest on a $ 20,000 loan he has taken . the bank charges him a quarterly compound rate of 5 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 20 % . here you find per quarter rate as ( 20 / 4 ) % = 5 % i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 5 % quarterly rate . so the bank charges 5 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 5 % of 20,000 = $ 4000 as our answer . but since , the interest is compounded , it will be a bit more than $ 2000 . option ( d ) looks correct ." | a = 100 * 100
b = 5 / 100
c = 1 + b
d = c ** 4
e = a * d
f = 100 * 100
g = e - f
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a ) 17 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | add(const_10, add(const_3, const_4)) | how many digits are in ( 8 Γ 10 ^ 10 ) ( 10 Γ 10 ^ 5 ) ? | the question simplfies to ( 8 Γ 10 ^ 10 ) ( 10 ^ 6 ) = > 8 * 10 ^ 16 = > will contain 16 zeros + 1 digit 8 = > 17 ans a | a = 3 + 4
b = 10 + a
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a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 16 : 9 | e | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(4, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 4 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 4 - - > e : l = 3 : 4 = 12 : 16 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 12 : 9 . thus , l : s = 16 : 9 . answer : e ." | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 4 * 4
f = d / e
g = c / f
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a ) rs . 120 , b ) rs . 160 , c ) rs . 270 , d ) rs . 300 , e ) none | c | subtract(subtract(multiply(divide(1530, const_10), const_2), const_12), const_12) | a sum of rs . 1530 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1530 = > 17 x / 12 = 1530 = > 1530 x 12 / 17 = rs . 1080 hence , b β s share = rs . ( 1080 / 4 ) = rs . 270 . answer c" | a = 1530 / 10
b = a * 2
c = b - 12
d = c - 12
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a ) 6.32 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 40))) | two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 40 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 / 40 = 3 / 19 hence , tank will be filled in 19 / 3 = 6.32 hour option ( a )" | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / 40
e = c - d
f = 1/(e)
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a ) 5 days , b ) 15 days , c ) 10 days , d ) 25 days , e ) 20 days | d | divide(multiply(15, 5), subtract(5, const_2)) | if 5 men undertook a piece of construction work and finished half the job in 15 days . if two men drop out , then the job will be completed in | that is , half the work done = 5 * 15 * 1 / 2 then , 5 * 15 * 1 / 2 = 3 * ? * 1 / 2 i . e . 5 * 15 = 3 * ? = 5 * 15 / 3 = 25 day d | a = 15 * 5
b = 5 - 2
c = a / b
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a ) 320 , b ) 340 , c ) 360 , d ) 380 , e ) 400 | d | multiply(divide(subtract(144, 72), const_3_6), 19) | two trains are moving in the same direction at 144 kmph and 72 kmph . the faster train crosses a man in the slower train in 19 seconds . find the length of the faster train ? | "relative speed = ( 144 - 72 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 19 sec = 19 * 20 = 380 m . the length of the faster train = 380 m . answer : d" | a = 144 - 72
b = a / const_3_6
c = b * 19
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a ) 2 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | e | divide(add(225, 43), 17) | a no . when divided by 225 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "225 + 43 = 268 / 17 = 13 ( remainder ) e" | a = 225 + 43
b = a / 17
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a ) 20 , b ) 120 , c ) 210 , d ) 15 , e ) 45 | b | choose(10, 3) | a child want to buy 3 new toys . if there are 10 toys that satisfy the child ' s criteria , in how many different ways can the child select the 3 new toys if the order of selection does n ' t matter . | "the number of ways of choosing 3 toys out of 10 are counted by = 10 c 3 ways = 120 answer : option b" | a = math.comb(10, 3)
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a ) 5 : 8 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 5 : 2 | e | divide(divide(subtract(96, 76), subtract(96, 68)), subtract(const_1, divide(subtract(96, 76), subtract(96, 68)))) | in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 76 per kg ? | "( 96 - 76 ) / ( 76 - 68 ) = 20 / 8 = 5 / 2 answer : e" | a = 96 - 76
b = 96 - 68
c = a / b
d = 96 - 76
e = 96 - 68
f = d / e
g = 1 - f
h = c / g
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['a ) 3 : 2', 'b ) 3 : 0', 'c ) 4 : 1', 'd ) 3 : 1', 'e ) 2 : 2'] | d | divide(3, divide(3, const_3)) | what is the ratio between perimeters of two squares one having 3 times the diagonal then the other ? | d = 3 d d = d a β 2 = 3 d a β 2 = d a = 3 d / β 2 a = d / β 2 = > 3 : 1 answer : d | a = 3 / 3
b = 3 / a
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a ) 12.5 % , b ) 10 % , c ) 15 % , d ) 17 % , e ) 19 % | b | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 72)), const_100), multiply(const_100, const_100)) | the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 72 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.72 x 1.25 = 0.9 decrease in price = ( 0.1 / 1 ) Γ 100 = 10 % b )" | a = 100 * 100
b = 100 + 25
c = b * 72
d = a - c
e = d * 100
f = 100 * 100
g = e / f
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a ) 10.9 sec , b ) 13.8 sec , c ) 53.8 sec , d ) 10.44 sec , e ) 10.4 sec | d | divide(add(140, 150), multiply(add(60, 40), const_0_2778)) | two trains 140 m and 150 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 150 = 290 m . required time = 290 * 9 / 250 = 10.44 sec . answer : d :" | a = 140 + 150
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 20,000 , b ) 21,200 , c ) 28,200 , d ) 13,500 , e ) none of these | d | divide(multiply(58000, divide(add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5))), add(add(add(multiply(20000, 5), multiply(subtract(20000, 10000), subtract(const_12, 5))), add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5)))), add(multiply(20000, 5), multiply(add(20000, 5000), subtract(const_12, 5)))))), add(const_60, 5)) | a , b and c start a business each investing 20000 . after 5 months a withdrew 10000 , b withdrew 15000 and c invests 5000 more . at the end of the year , a total profit of 58000 was recorded . find the share of b . | ratio of the capitals of a , b and c = 20000 Γ£ β 5 + 10000 Γ£ β 7 : 20000 Γ£ β 5 + 5000 Γ£ β 7 : 20000 Γ£ β 5 + 25000 Γ£ β 7 = 170000 : 135000 : 275000 = 170 : 135 : 275 . b Γ’ β¬ β’ s share = ( 58000 Γ£ β 135 Γ’ Β β 580 ) = 13500 answer d | a = 20000 * 5
b = 20000 - 15000
c = 12 - 5
d = b * c
e = a + d
f = 20000 * 5
g = 20000 - 10000
h = 12 - 5
i = g * h
j = f + i
k = 20000 * 5
l = 20000 - 15000
m = 12 - 5
n = l * m
o = k + n
p = j + o
q = 20000 * 5
r = 20000 + 5000
s = 12 - 5
t = r * s
u = q + t
v = p + u
w = e / v
x = 58000 * w
y = const_60 + 5
z = x / y
|
a ) 0.002 % , b ) 0.02 % , c ) 0.2 % , d ) 12 % , e ) 20 % | d | multiply(divide(multiply(0.06, 20), 10), const_100) | a glass was filled with 10 ounces of water , and 0.06 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ? | "we are given that 0.06 ounces of water evaporated each day . furthermore , we know that this process happened over a 20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply 0.06 by 20 . this gives us : 0.06 x 20 = 1.2 ounces finally , we are asked for β what percent β of the original amount of water evaporated during this period . to determine this percentage , we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated / original amount ) x 100 % ( 1.2 / 10 ) x 100 % ( 12 / 100 ) x 100 % = 12 % answer d" | a = 0 * 6
b = a / 10
c = b * 100
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | subtract(multiply(25, const_2.0), add(multiply(subtract(subtract(25, add(add(multiply(11, 1), 6), 3)), 1), 3), add(multiply(11, 1), multiply(6, 3)))) | in a class of 25 students , 3 students did not borrow any books from the library , 11 students each borrowed 1 book , 6 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the total number of books the students borrowed is 25 * 2 = 50 . the students who borrowed zero , one , or two books borrowed 11 * 1 + 6 * 2 = 23 books . the 5 students who borrowed at least three books borrowed 50 - 23 = 27 books . if 4 of these students borrowed exactly 3 books , then the maximum that one student could have borrowed is 27 - 12 = 15 books . the answer is d ." | a = 25 * 2
b = 11 * 1
c = b + 6
d = c + 3
e = 25 - d
f = e - 1
g = f * 3
h = 11 * 1
i = 6 * 3
j = h + i
k = g + j
l = a - k
|
a ) 8 , b ) 6 , c ) 5 , d ) 4 , e ) 1 | b | divide(multiply(multiply(6, const_3), const_3), multiply(const_3, const_3)) | if two painters can complete two rooms in two hours , how many painters would it take to do 18 rooms in 6 hours ? | "explanation : two painters can complete two rooms in two hours . so 18 rooms can be painted in 6 hrs by 6 painters answer : b ) 6 painters" | a = 6 * 3
b = a * 3
c = 3 * 3
d = b / c
|
a ) 878 , b ) 879 , c ) 880 , d ) 881 , e ) 882 | e | multiply(subtract(14, multiply(7, const_2)), multiply(21, 17)) | how much space , in cubic units , is left vacant when maximum number of 7 x 7 x 7 cubes are fitted in a rectangular box measuring 14 x 21 x 17 ? | "no of cubes that can be accommodated in box = ( 14 * 21 * 17 ) / ( 7 * 7 * 7 ) 12 * 16 in numerator can be perfectly divided by 7 * 7 in denominator . side with length 17 ca n ' t be perfectly divided by 7 and hence is the limiting factor . closet multiple of 7 less that 17 is 14 . so vacant area in cube = = 14 * 21 * ( 17 - 14 ) = 14 * 21 * 3 = 882 ans - e" | a = 7 * 2
b = 14 - a
c = 21 * 17
d = b * c
|
a ) 30 m , b ) 40 m , c ) 35 m , d ) 38 m , e ) 50 m | a | multiply(divide(72, multiply(divide(30, const_100), divide(subtract(const_100, 20), const_100))), multiply(divide(50, const_100), divide(20, const_100))) | according to a recent survey report issued by the commerce ministry , government of india , 30 % of the total fdi goes to gujarat and 20 % of this goes to rural areas . if the fdi in gujarat , which goes to urban areas , is $ 72 m , then find the size of fdi in rural andhra pradesh , which attracts 50 % of the fdi that comes to andhra pradesh , which accounts for 20 % of the total fdi ? | 30 % of total fdi given to gujarat is $ 90 m . so , 100 % is $ 300 m . 20 % of 300 m = 60 m . 50 % of 60 m = 30 m . answer : a | a = 30 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 72 / d
f = 50 / 100
g = 20 / 100
h = f * g
i = e * h
|
a ) 9 : 10 , b ) 17 : 19 , c ) 23 : 27 , d ) 16 : 19 , e ) 15 : 23 | d | sqrt(divide(1792, 2527)) | triangle atriangle b are similar triangles with areas 1792 units square and 2527 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1792 / 2527 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 7 ) / ( 19 * 19 * 7 ) ( x ^ 2 / y ^ 2 ) = 16 ^ 2 / 19 ^ 2 x / y = 16 / 19 ans = d" | a = 1792 / 2527
b = math.sqrt(a)
|
a ) $ 310 , b ) $ 395 , c ) $ 484 , d ) $ 512 , e ) $ 542 | c | divide(1210, add(divide(divide(2, 5), divide(4, 15)), const_1)) | a and b together have $ 1210 . if 4 / 15 of a ' s amount is equal to 2 / 5 of b ' s amount , how much amount does b have ? | 4 / 15 a = 2 / 5 b a = 2 / 5 * 15 / 4 b a = 3 / 2 b a / b = 3 / 2 a : b = 3 : 2 b ' s share = 1210 * 2 / 5 = $ 484 answer is c | a = 2 / 5
b = 4 / 15
c = a / b
d = c + 1
e = 1210 / d
|
a ) 87 , b ) 98 , c ) 12 , d ) 27 , e ) 28 | c | add(divide(divide(square_edge_by_area(784), const_2), divide(add(const_3, const_4), const_3)), const_2) | there are two circles of different radii . the are of a square is 784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle . ? | "let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a 2 = 784 = ( 4 ) ( 196 ) = ( 22 ) . ( 142 ) a = ( 2 ) ( 14 ) = 28 a = 2 l , l = a / 2 = 14 l = ( 7 / 3 ) s therefore s = ( 3 / 7 ) ( l ) = 6 circumference of the smaller circle = 2 β s = 12 β cm . answer : c" | a = square_edge_by_area / (
b = a / 2
c = 3 + 4
d = c / 3
e = b + d
|
a ) 1.6 km , b ) 2 km , c ) 3.6 km , d ) 4.8 km , e ) none of these | d | multiply(divide(18, const_60), add(12, 4)) | the speed of a boat in still water is 12 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 18 minutes is | "explanation : speed downstreams = ( 12 + 4 ) kmph = 16 kmph . distance travelled = ( 16 x 18 / 60 ) km = 4.8 km option d" | a = 18 / const_60
b = 12 + 4
c = a * b
|
a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | b | subtract(divide(add(multiply(add(const_4, const_3), 100), multiply(subtract(const_10, const_1), 20)), 20), add(2, 2)) | a gambler bought $ 3000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 16 : x + x - 2 = 16 - - > x = 9 : 9 $ 20 chips were lost and 9 - 2 = 7 $ 100 chips were lost . total worth of chips lost is 9 * 20 + 7 * 100 = $ 880 , so the gambler kept $ 3,000 - $ 880 = $ 2,120 . answer : b . | a = 4 + 3
b = a * 100
c = 10 - 1
d = c * 20
e = b + d
f = e / 20
g = 2 + 2
h = f - g
|
a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) none | d | divide(add(multiply(divide(const_3, const_2), 3), 3), subtract(divide(const_3, const_2), 1)) | a boat covers a certain distance downstream in 1 hour , while it comes back in 1 Β½ hours . if the speed of the stream be 3 kmph , what is the speed of the boat in still water ? | sol . let the speed of the boat in still water be x kmph . then , speed downstream = ( x + 3 ) kmph , speed upstream = ( x - 3 ) kmph . β΄ ( x + 3 ) * 1 = ( x - 3 ) * 3 / 2 β 2 x + 6 = 3 x - 9 β x = 15 kmph . answer d | a = 3 / 2
b = a * 3
c = b + 3
d = 3 / 2
e = d - 1
f = c / e
|
a ) 0.52 , b ) 0.29 , c ) 0.48 , d ) 0.64 , e ) 0.46 | b | add(multiply(0.3, 0.8), multiply(0.1, 0.5)) | simplify : 0.3 * 0.8 + 0.1 * 0.5 | "given exp . = 0.3 * 0.8 + ( 0.1 * 0.5 ) = 0.24 + 0.05 = 0.29 answer is b ." | a = 0 * 3
b = 0 * 1
c = a + b
|
a ) 65 , b ) 70 , c ) 75 , d ) 80 , e ) 85 | d | multiply(inverse(add(divide(const_1, 4), divide(const_1, 2))), const_60) | working alone , pump a can empty a pool in 4 hours . working alone , pump b can empty the same pool in 2 hours . working together , how many minutes will it take pump a and pump b to empty the pool ? | "pump a can empty ( 1 / 4 ) of the pool per hour . pump b can empty ( 1 / 2 ) of the pool per hour . together the pumps can empty 1 / 4 + 1 / 2 = 3 / 4 of the pool per hour . 1 pool / ( 3 / 4 ) pool per hour = 4 / 3 hours = 80 minutes . the answer is d ." | a = 1 / 4
b = 1 / 2
c = a + b
d = 1/(c)
e = d * const_60
|
a ) 1977 , b ) 1893 , c ) 2700 , d ) 1900 , e ) 1278 | c | subtract(divide(subtract(multiply(2000, 62), multiply(2000, 15)), 20), 2000) | a garrison of 2000 men has provisions for 62 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 - - - - 62 2000 - - - - 47 x - - - - - 20 x * 20 = 2000 * 47 x = 4700 2000 - - - - - - - 2700 answer : c" | a = 2000 * 62
b = 2000 * 15
c = a - b
d = c / 20
e = d - 2000
|
['a ) β ( 2 / 3 ) Γ x', 'b ) 2 / β 3 Γ x', 'c ) β 3 x', 'd ) β ( 8 / 3 ) Γ x', 'e ) 2 β 3 Γ x'] | d | divide(multiply(power(add(const_1, const_1), inverse(const_2)), const_2), power(const_3, inverse(const_2))) | the diagonal of a square is the height in an equilateral triangle . if the side of the square is x , what is the side of the equilateral triangle in terms of x ? | the diagonal of the square x * sqrt ( 2 ) if that ' s the altitude of the equilateral triangle , the formula for which is side * sqrt ( 3 ) / 2 side is 2 sqrt ( 2 ) * x / sqrt ( 3 ) answer : ( option d ) | a = 1 + 1
b = 1/(2)
c = a ** b
d = c * 2
e = 1/(2)
f = 3 ** e
g = d / f
|
a ) 2 / 3 , b ) 3 / 5 , c ) 2 / 7 , d ) 3 / 7 , e ) 4 / 9 | a | divide(divide(4, 17), divide(6, 17)) | if p ( a ) = 6 / 17 , p ( b ) = 5 / 17 , and p ( a βͺ b ) = 4 / 17 find p ( b | a ) ? | "p ( b | a ) = p ( a βͺ b ) / p ( a ) p ( b | a ) = ( 4 / 17 ) / ( 6 / 17 ) = 4 / 6 = 2 / 3 . a" | a = 4 / 17
b = 6 / 17
c = a / b
|
a ) 152.5 , b ) 143.75 , c ) 155.5 , d ) 165.5 , e ) 117.5 | b | multiply(divide(multiply(add(divide(15, const_100), const_1), 25), 20), const_100) | last year elaine spent 20 % of her annual earnings on rent . this year she earned 15 % more than last year and she spent 25 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ? | for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 20 of this on rent . this year she earned 15 % more , or $ 115 . she would ' ve spent 25 % of this on rent , or $ 28.75 do $ 34.5 / $ 20 this will give you 143.75 % b is the correct answer . | a = 15 / 100
b = a + 1
c = b * 25
d = c / 20
e = d * 100
|
a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 1000 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 20,000 is invested at the same interest rate , what is the interest from the investment ? | - > 250 / 5,000 = 5 % and 20,000 * 5 % = 1000 . thus , d is the answer . | a = 2 * 100
b = a * 100
c = 250 * b
d = 2 * 100
e = d * 100
f = e / 4
g = c / f
|
a ) 800 , b ) 900 , c ) 950 , d ) 1000 , e ) 1050 | b | multiply(divide(multiply(180, const_1000), const_3600), 18) | a train running at the speed of 180 km / hr crosses a pole in 18 seconds . what is the length of the train ? | "speed = ( 180 x ( 5 / 18 ) m / sec = ( 50 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 50 ) x 18 ) m = 900 m b" | a = 180 * 1000
b = a / 3600
c = b * 18
|
a ) 165 , b ) 170 , c ) 175 , d ) 180 , e ) 225 | e | multiply(divide(const_100, 8), 18) | a 18 % stock yielding 8 % is quoted at ? | "assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 18 , money invested = 100 Γ 18 / 8 = rs . 225 ie , market value of the stock = rs . 175 answer is e ." | a = 100 / 8
b = a * 18
|
a ) 50 days , b ) 110.6 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(145, const_100), 60)) | p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 145 days , in how many days can r alone complete the remaining work ? | p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 395 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 145 days - - > 145 / 250 work done = > 58 % 28 % work left . . . r alone will take 28 % * 395 = 110.6 days answer is ( b ) | a = 4 * 50
b = 145 / 100
c = b * 60
d = a - c
|
a ) 160 , b ) 72 , c ) 112 , d ) 128 , e ) 142 | a | multiply(2, divide(180, add(2, 16))) | water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 180 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is 180 18 x = 180 x = 10 since x is 10 , we know that there are 16 x 10 = 160 grams of oxygen in 180 grams of water . answer a ." | a = 2 + 16
b = 180 / a
c = 2 * b
|
a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | b | divide(subtract(40, 26), subtract(40, 19)) | a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containing 19 % alcohol and now the percentage of alcohol was found to be 26 % . the quantity of whisky replaced is : | "explanation : by the rule of alligation , we have : so , ratio of 1 st and 2 nd quantities = 7 : 14 = 1 : 2 required quantity replaced = 2 / 3 . answer is b" | a = 40 - 26
b = 40 - 19
c = a / b
|
a ) 21 kmph , b ) 22 kmph , c ) 20 kmph , d ) 23 kmph , e ) 24 kmph | c | multiply(divide(20, const_60), 60) | the speed of a train is 60 kmph . what is the distance covered by it in 20 minutes ? | "60 * 20 / 60 = 20 kmph answer : c" | a = 20 / const_60
b = a * 60
|
a ) 15 , b ) 19 , c ) 18 , d ) 22 , e ) none | c | divide(add(add(9, const_4), subtract(26, const_4)), const_2) | find the average of all the numbers between 9 and 26 which are divisible by 5 . | "sol . average = ( 10 + 15 + 20 + 25 / 4 ) = 70 / 4 = 17.5 answer c" | a = 9 + 4
b = 26 - 4
c = a + b
d = c / 2
|
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