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a ) 4 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | a | divide(8, subtract(4, 2)) | a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 8 km ? | "m = 4 s = 2 us = 4 - 2 = 2 d = 8 t = 8 / 2 = 4 answer : a" | a = 4 - 2
b = 8 / a
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a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(2, const_4) | if a and b are both odd prime numbers and a < b , then how many different positive integer factors does 2 ab have ? | a and b are both odd prime numbers - it means either a or b is not 2 and since prime numbers have only two factors - 1 and the number itself a and b each will have ( 1 + 1 ) = 2 factors hence 2 ab will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer | a = 2 * 4
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a ) 2000 , b ) 2777 , c ) 2688 , d ) 1000 , e ) 2871 | a | divide(subtract(130, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100))) | rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 130 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 130 x = 2000 answer : a" | a = 6 * 2500
b = a / 100
c = 130 - b
d = 5 / 100
e = 6 / 100
f = d - e
g = c / f
|
a ) 360 , b ) 420 , c ) 510 , d ) 525 , e ) 280 | d | lcm(lcm(15, 25), 35) | find the lowest common multiple of 15 , 25 and 35 . | "lcm = 3 * 5 * 5 * 7 = 525 . answer is d" | a = math.lcm(15, 25)
b = math.lcm(a, 35)
|
a ) 1 km , b ) 1500 mts , c ) 1600 mts , d ) 2 km , e ) 1250 mts | b | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 5), divide(const_1, 10)), divide(const_1, 15))), const_3), const_1000) | a person travels equal distances with speeds of 5 km / hr , 10 km / hr , 15 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 5 ) + ( x / 10 ) + ( x / 15 ) = 11 / 60 x = 0.5 total distance = 3 * 0.5 = 1.5 km = 1500 meters correct option is b" | a = 11 / const_60
b = 1 / 5
c = 1 / 10
d = b + c
e = 1 / 15
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | b | multiply(add(multiply(add(multiply(add(const_1, 3), 2), 2), 2), const_1), 2) | a charitable woman gave one penny more than half pennies in her purse to a farmer . afterwards she came across a beggar whom she gave 2 pennies more than half of what she had in her purse . in the end she gave 3 pennies more than half of what she had in her purse and left with a penny with her to a boy . find what she had in the beginning ? | she gave 1 penny more than the half in the purse i . e ( x / 2 + 1 ) now she is left with : : ( x - ( x / 2 + 1 ) ) = ( x - 2 ) / 2 now to the beggar : : she gave 2 pennies more than the half in the purse i . e ( x - 2 ) / 4 + 2 and she is left with : : ( ( x - 2 ) / 2 ) - ( ( x - 2 ) / 4 + 2 ) = ( x - 10 ) / 4 now at last sh gave 3 pennies more then the half she was left with i . e ( ( x - 10 ) / 8 + 3 ) = a so she is left with ( x - 10 ) / 4 - a = ( x - 34 ) / 8 now acc . to quest ( x - 34 ) / 8 = 1 hence x = 42 answer : b | a = 1 + 3
b = a * 2
c = b + 2
d = c * 2
e = d + 1
f = e * 2
|
a ) 15 , b ) 26 , c ) 37 , d ) 42 , e ) 60 | b | add(add(divide(divide(multiply(multiply(2, 5), 6), const_2), 2), divide(divide(multiply(multiply(2, 5), 6), const_2), 5)), divide(divide(multiply(multiply(2, 5), 6), const_2), 6)) | if x , y , and z are positive integers , and 2 x = 5 y = 6 z , then the least possible value of x + y + z is | "take lcm of 2,5 and 6 = 30 now 2 x = 30 = > x = 15 5 y = 30 = > y = 6 6 z = 30 = > z = 5 15 + 6 + 5 = 26 . option b ." | a = 2 * 5
b = a * 6
c = b / 2
d = c / 2
e = 2 * 5
f = e * 6
g = f / 2
h = g / 5
i = d + h
j = 2 * 5
k = j * 6
l = k / 2
m = l / 6
n = i + m
|
['a ) w ^ 2 + pw + x = 0', 'b ) w ^ 2 - pw + 2 x = 0', 'c ) 2 w ^ 2 + pw + 2 x = 0', 'd ) 2 w ^ 2 - pw - 2 x = 0', 'e ) 2 w ^ 2 - pw + 2 x = 0'] | e | add(subtract(multiply(const_2, const_1), add(multiply(const_2, power(const_1, const_2)), multiply(const_2, power(const_1, const_2)))), multiply(const_2, power(const_1, const_2))) | a rectangular circuit board is designed to have width w inches , perimeter p inches , and area x square inches . which of the following equations must be true ? | notice that we can discard options a , and c right away . the sum of 3 positive values can not be 0 . now , assume : width = w = 1 inchand length = 1 inch ; perimeter = p = 4 inches ; area = x = 1 square inches . plug the values of w , p , and x into the answer choices : only for e 2 w ^ 2 - pw + 2 x = 2 - 4 + 2 = 0 . answer : e . | a = 2 * 1
b = 1 ** 2
c = 2 * b
d = 1 ** 2
e = 2 * d
f = c + e
g = a - f
h = 1 ** 2
i = 2 * h
j = g + i
|
a ) 81000 , b ) 85000 , c ) 75000 , d ) 72000 , e ) 77000 | d | subtract(add(multiply(power(const_10, const_4), const_10), 80000), divide(multiply(divide(add(multiply(power(const_10, const_4), const_10), 80000), const_2), multiply(const_2, const_3)), add(const_2, const_3))) | one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 80000 as total savings , how much has he saved in public provident fund ? | let savings in national savings certificate = x and savings in public provident fund = ( 180000 - x ) 1 / 3 x = 1 / 2 ( 180000 β x ) β 2 x = 3 ( 180000 β x ) β 2 x = 540000 β 3 x β 5 x = 540000 β x = 540000 / 5 = 108000 savings in public provident fund = ( 180000 - 108000 ) = 72000 answer is d . | a = 10 ** 4
b = a * 10
c = b + 80000
d = 10 ** 4
e = d * 10
f = e + 80000
g = f / 2
h = 2 * 3
i = g * h
j = 2 + 3
k = i / j
l = c - k
|
a ) 1 : 2 , b ) 3 : 4 , c ) 5 : 6 , d ) 7 : 8 , e ) 9 : 10 | e | divide(add(1, divide(7, 2)), add(divide(7, 2), divide(3, 2))) | the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 2 type b widgets , and the weight of 3 type b widgets is equal to the weight of 7 type c widgets . what is the ratio of the total weight of 1 type a widget and 1 type b widget , to the total weight of 1 type b widget and 1 type c widget ? | 3 b = 7 c and so b = 7 c / 3 7 a = 2 b and so a = 2 b / 7 = 2 c / 3 a + b = 2 c / 3 + 7 c / 3 = 9 c / 3 b + c = 7 c / 3 + c = 10 c / 3 the ratio of a + b : b + c = 9 : 10 the answer is e . | a = 7 / 2
b = 1 + a
c = 7 / 2
d = 3 / 2
e = c + d
f = b / e
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1170) | what least number must be added to 1170 , so that the sum is completely divisible by 23 ? | "51 * 23 = 1173 1173 - 1170 = 3 answer : a" | a = 4 * 10
b = a + 2
c = b * 23
d = c - 1170
|
a ) 149 , b ) 147 , c ) 101 , d ) 143 , e ) 141 | c | add(multiply(multiply(divide(50, const_100), 20), multiply(divide(50, const_100), 20)), divide(subtract(500, 20), 20)) | in a 500 member association consisting of men and women , exactly 20 % of men and exactly 50 % women are homeowners . what is the least number of members who are homeowners ? | "solution simple out of 500 20 % are male i . e 100 and 50 % are female i . e 250 , so total homeowner is 350 . now min number homeowner is 100 and max is 250 so question ask us to find least and 101 has least value among all option . so ans is 101 . ans : c" | a = 50 / 100
b = a * 20
c = 50 / 100
d = c * 20
e = b * d
f = 500 - 20
g = f / 20
h = e + g
|
a ) 5.6 sec , b ) 8.9 sec , c ) 10 sec , d ) 12.6 sec , e ) 15 sec | c | divide(add(140, 160), multiply(add(60, 48), const_0_2778)) | two trains 140 m and 160 m long run at the speed of 60 kmph and 48 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 48 = 108 kmph * 5 / 18 = 30 m / s distance covered in crossing each other = 140 + 160 = 300 m required time = 300 * 1 / 30 = 10 sec answer is c" | a = 140 + 160
b = 60 + 48
c = b * const_0_2778
d = a / c
|
a ) 25 , b ) 32 , c ) 35 , d ) 30 , e ) 45 | a | divide(add(408, 192), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | there are 408 boys and 192 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | "explanation : hcf ( 408 , 192 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 192 / 24 = 17 + 8 = 25 answer : a" | a = 408 + 192
b = 2 * 2
c = b * 2
d = c * 3
e = a / d
|
a ) 18 , b ) 128 , c ) 175 , d ) 216 , e ) 250 | b | add(multiply(divide(360, 4), 1), divide(subtract(360, multiply(divide(360, 4), 1)), 7)) | joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 4 of all the paint . during the second week , he uses 1 / 7 of the remaining paint . how many gallons of paint has joe used ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 4 ) * 360 = 90 gallons . remaning paint = 270 gallons paint used in the second week = ( 1 / 7 ) * 270 = 38 gallons total paint used = 128 gallons . option b" | a = 360 / 4
b = a * 1
c = 360 / 4
d = c * 1
e = 360 - d
f = e / 7
g = b + f
|
a ) 288 , b ) 500 , c ) 502 , d ) 540 , e ) 262 | c | subtract(subtract(540, divide(multiply(540, 5), const_100)), divide(multiply(subtract(540, divide(multiply(540, 5), const_100)), 2), const_100)) | the sale price sarees listed for rs . 540 after successive discount is 5 % and 2 % is ? | "540 * ( 95 / 100 ) * ( 98 / 100 ) = 502 answer : c" | a = 540 * 5
b = a / 100
c = 540 - b
d = 540 * 5
e = d / 100
f = 540 - e
g = f * 2
h = g / 100
i = c - h
|
a ) 2 / 21 , b ) 3 / 41 , c ) 3 / 28 , d ) 3 / 31 , e ) 5 / 32 | c | divide(choose(3, 2), choose(add(add(3, 2), 3), 2)) | a bag contains 3 red , 2 blue and 3 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 3 c 2 / 8 c 2 = 6 c 213 c 2 = 3 / 28 c" | a = math.comb(3, 2)
b = 3 + 2
c = b + 3
d = math.comb(c, 2)
e = a / d
|
a ) 391 , b ) 417 , c ) 410 , d ) 423 , e ) 445 | b | divide(480, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 480 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 480 j = 480 / 1.15 j = 48,000 / 115 = 417 thus b is the correct answer ." | a = 15 / 100
b = 1 + a
c = 480 / b
|
a ) 9 , b ) 12 , c ) 15 , d ) 16 , e ) 26.6 | e | subtract(multiply(divide(multiply(40, divide(5, add(5, 3))), divide(3, add(5, 3))), divide(5, add(5, 3))), subtract(40, multiply(40, divide(5, add(5, 3))))) | at deluxe paint store , fuchsia paint is made by mixing 5 parts of red paint with 3 parts of blue paint . mauve paint is made by mixing 3 parts of red paint with 5 parts blue paint . how many liters of blue paint must be added to 40 liters of fuchsia to change it to mauve paint ? | "in 40 liters , red = 5 / 8 * 40 = 25 and blue = 15 so , 25 / ( 15 + x ) = 3 / 5 or , x = 26.6 ( answer e )" | a = 5 + 3
b = 5 / a
c = 40 * b
d = 5 + 3
e = 3 / d
f = c / e
g = 5 + 3
h = 5 / g
i = f * h
j = 5 + 3
k = 5 / j
l = 40 * k
m = 40 - l
n = i - m
|
a ) 3170 , b ) 3172 , c ) 3174 , d ) 3168 , e ) 3167 | d | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(4, 4)), multiply(const_4, const_2))), const_10) | how many 4 digits number contain the number 2 . | "total no . of 4 digit nos are 9000 ( btwn 1000 and 9999 ) . consider the case where the nos contain no 2 in it . so units place can be filled in 9 ways ( 0 , 1,3 , 4,5 , 6,7 , 8,9 ) smlry , tenths place in 9 ways hunderedths place in 9 ways thousands place can be filled in 8 ways only ( because 0 can not come ) so total nos are ( 9000 - ( 8 * 9 * 9 * 9 ) ) = 3168 answer : d" | a = 1000 - 10
b = 4 * 4
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) a ) 25 , b ) b ) 34 , c ) c ) 50 , d ) d ) 40 , e ) e ) 100 | d | divide(divide(multiply(320, 6), 12), const_4) | according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 320 6 ounces servings of orange juice ? | its d . total juice rquired = 320 * 6 = 1920 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1920 / 48 = 40 . answer d | a = 320 * 6
b = a / 12
c = b / 4
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a ) 6 , b ) 12 , c ) 24 , d ) 120 , e ) none | c | factorial(4) | the largest natural number which exactly divides the product of any 4 consecutive natural numbers is | sol . required number = 1 Γ 2 Γ 3 Γ 4 = 24 . answer c | a = math.factorial(4)
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a ) 100 % , b ) 500 % , c ) 300 % , d ) 600 % , e ) 800 % | b | multiply(const_100, divide(const_2, const_2)) | the area of a circle is increased by 1500 % . by what percent has the diameter of the circle increased ? | "the area of the circle is increased by 1500 % , thus the area is increased 16 times . the area of a circle it proportional to the square of the diameter ( area = Ο d ^ 2 / 4 ) , therefore the diameter must increase 4 times ( diameter increase 4 times = area increase 16 times ) , which is increase by 500 % . answer : b ." | a = 2 / 2
b = 100 * a
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a ) 7 days , b ) 4 days , c ) 6 days , d ) 8 days , e ) 9 days | b | divide(const_1, add(multiply(12, divide(divide(const_1, 6), 24)), multiply(16, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 16 boys and 12 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 16 boys + 12 girls = 16 Γ£ β 6 / 4 + 12 = 24 + 12 = 36 girls 4 days to complete the work answer : option b" | a = 1 / 6
b = a / 24
c = 12 * b
d = 1 / 6
e = d / 16
f = 16 * e
g = c + f
h = 1 / g
|
a ) 3327 , b ) 3237 , c ) 3927 , d ) 2337 , e ) none of these | c | subtract(13800, 9873) | 9873 + x = 13800 , then x is ? | "answer x = 13800 - 9873 = 3927 option : c" | a = 13800 - 9873
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a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 3 | e | subtract(add(add(20, 18), 5), 40) | in a sports club with 40 members , 20 play badminton and 18 play tennis and 5 do not play either . how many members play both badminton and tennis ? | "20 + 18 = 38 but where as total number is 40 - 5 = 35 therefore answer is 38 - 35 = 3 hence answer is e" | a = 20 + 18
b = a + 5
c = b - 40
|
a ) 12 % , b ) 16 % , c ) 19 % , d ) 50 % , e ) 60 % | d | subtract(const_100, divide(multiply(300, const_100), 600)) | an article is bought for rs . 600 and sold for rs . 300 , find the loss percent ? | "600 - - - - 300 100 - - - - ? = > 50 % answer : d" | a = 300 * 100
b = a / 600
c = 100 - b
|
a ) 10 , b ) 20 , c ) 30 , d ) 50 , e ) 40 | b | sqrt(divide(10, add(power(2, 1), add(power(2, 1), power(const_2.0, const_2))))) | the ratio of 2 numbers is 1 : 2 and their h . c . f . is 10 . their l . c . m . is ? | "let the numbers be x and 2 x their h . c . f . = 10 so the numbers are 1 * 10 , 2 * 10 = 10,20 l . c . m . = 20 answer is b" | a = 2 ** 1
b = 2 ** 1
c = 2 ** 0
d = b + c
e = a + d
f = 10 / e
g = math.sqrt(f)
|
a ) 286 m , b ) 250 m , c ) 277 m , d ) 278 m , e ) 267 m | b | subtract(multiply(speed(300, 18), 33), 300) | a 300 m long train crosses a platform in 33 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 33 = 50 / 3 = > x = 250 m . answer : b" | a = speed * (
b = a - 33
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a ) 4 , b ) 7 , c ) 10 , d ) 15 , e ) 21 | e | divide(multiply(14, const_3), const_2) | youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 14 minutes more to walk to work than to ride his bike to work , then x equals ? | please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 840 ( which is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 840 - - - - > 40 x = 840 - > x = 21 . option e | a = 14 * 3
b = a / 2
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a ) 337 , b ) 429 , c ) 266 , d ) 288 , e ) 211 | b | divide(60, divide(multiply(4, add(3, divide(1, 2))), const_100)) | what sum of money will produce rs . 60 as simple interest in 4 years at 3 1 / 2 percent ? | "60 = ( p * 4 * 7 / 2 ) / 100 p = 429 answer : b" | a = 1 / 2
b = 3 + a
c = 4 * b
d = c / 100
e = 60 / d
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a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 34 , e ) 6 : 24 | d | subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 45))) | a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 45 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ? | "c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 6 : 34 pm . the correct answer is d ." | a = 6 * 10
b = const_60 + 15
c = b / const_60
d = const_60 - 20
e = d / const_60
f = c * e
g = const_60 + 20
h = g / const_60
i = f * h
j = const_60 - 45
k = i * j
l = a - k
|
a ) 141 , b ) 142 , c ) 143 , d ) 144 , e ) 145 | c | subtract(speed(speed(140, 3), const_0_2778), 25) | an woman sitting in a train which is travelling at 25 kmph observes that a goods train travelling in a opposite direction , takes 3 seconds to pass him . if the goods train is 140 m long , find its speed . | "relative speed = ( 140 / 3 ) m / s = ( 140 / 3 ) * ( 18 / 5 ) = 168 kmph speed of goods train = 168 - 25 = 143 kmph answer is c" | a = speed - (
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a ) 8.4 , b ) 8.6 , c ) 8.8 , d ) 9 , e ) 9.2 | d | divide(add(18, 18), subtract(7.3, add(divide(18, 12), divide(18, 10)))) | on a trip , a cyclist averaged 12 miles per hour for the first 18 miles and 10 miles per hour for the remaining 18 miles . if the cyclist returned immediately via the same route and took a total of 7.3 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 36 miles was 18 / 12 + 18 / 10 = 1.5 + 1.8 = 3.3 hours . the average speed for the return trip was 36 miles / 4 hours = 9 mph . the answer is d ." | a = 18 + 18
b = 18 / 12
c = 18 / 10
d = b + c
e = 7 - 3
f = a / e
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a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | e | add(multiply(power(divide(128, multiply(1, multiply(1, const_2))), inverse(const_3)), multiply(1, const_2)), multiply(1, const_2)) | there is a rectangular prism made of 1 in cubes that has been covered in tin foil . there are exactly 128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these 128 cubes is twice the length and twice the height , what is the measure q in inches of the width of the foil covered prism ? | "if the width is w , then length and height would be w / 2 . so , w * w / 2 * w / 2 = 128 = > w ^ 3 = ( 2 ^ 3 ) * 64 = ( 2 ^ 3 ) * ( 4 ^ 3 ) = > w = 2 * 4 = 8 in . along the width of the cuboid , 8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = 8 + 2 = q = 10 ans e ." | a = 1 * 2
b = 1 * a
c = 128 / b
d = 1/(3)
e = c ** d
f = 1 * 2
g = e * f
h = 1 * 2
i = g + h
|
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(30, multiply(add(1, const_2), const_2)) | in covering a distance of 30 km , arun takes 2 hours more than anil . if arun doubles his speed , then he would take 1 hour less than anil . what is arun ' s speed ? | "let the speed of arun = x kmph and the speed of anil = y kmph distance = 30 km we know that distancespeed = timehence 30 x β 30 y = 2 . . . . . . . . . . . ( equation 1 ) 30 y β 302 x = 1 . . . . . . . . . . . ( equation 2 ) equation 1 + equation 2 β 30 x β 302 x = 3 β 302 x = 3 β 15 x = 3 β 5 x = 1 β x = 5 hence arun ' s speed = 5 kmph answer : b" | a = 1 + 2
b = a * 2
c = 30 / b
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a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | e | multiply(divide(multiply(multiply(150, divide(150, 40)), const_2), 150), const_2) | the racing magic takes 150 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | "time taken by racing magic to make one circle = 150 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 150 seconds = 450 seconds time taken for them to meet at the starting point for the second time = 2 * 450 = 900 seconds = 15 mins answer e" | a = 150 / 40
b = 150 * a
c = b * 2
d = c / 150
e = d * 2
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a ) 80 cm , b ) 82 cm , c ) 90 cm , d ) 94.8 cm , e ) 95 cm | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 144 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 144 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 52.8 + 42 = 94.8 cm answer : d" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 210 , b ) 252 , c ) 280 , d ) 300 , e ) 420 | d | divide(multiply(multiply(25, 42), 60), multiply(multiply(7, 6), 5)) | a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | maximum number of soap boxes that can be placed in each carton = volume of each carton / volume of each box = ( 25 * 42 * 60 ) / ( 7 * 6 * 5 ) = 300 answer : d | a = 25 * 42
b = a * 60
c = 7 * 6
d = c * 5
e = b / d
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a ) 456 , b ) 789 , c ) 991 , d ) 1020 , e ) 1100 | c | add(divide(subtract(10000, 100), 10), const_1) | how many multiples of 10 are there between 100 and 10000 ( both are inclusive ) ? | "the answer is ( 10000 - 100 ) / 10 + 1 = 991 answer is c" | a = 10000 - 100
b = a / 10
c = b + 1
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a ) 43.1543 , b ) 43.1818 , c ) 43.45128 , d ) 43.41288 , e ) 43.41222 | b | multiply(divide(25, 12), 20) | there are 20 poles with a constant distance between each pole . a car takes 25 second to reach the 12 th pole . how much will it take to reach the last pole . | "assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 25 seconds 1 pole ( 25 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 25 / 11 ) seconds = 43.1818 seconds answer : b" | a = 25 / 12
b = a * 20
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a ) 5 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | a | multiply(add(divide(const_1, 25), divide(const_1, 20)), const_100) | if x > 0 , x / 25 + x / 20 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 25 + 100 / 20 = 4 + 5 = 9 9 is 9 % of 100 = a ." | a = 1 / 25
b = 1 / 20
c = a + b
d = c * 100
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a ) 45 , 63 , b ) 50 , 70 , c ) 55 , 77 , d ) 60 , 84 , e ) 65 , 91 | b | subtract(multiply(multiply(subtract(multiply(9, 25), multiply(5, 25)), 5), divide(7, 5)), multiply(subtract(multiply(9, 25), multiply(5, 25)), 5)) | two numbers are in the ratio of 5 : 7 . if 25 is subtracted from each number , they are in the ratio of 5 : 9 . what are the two numbers ? | "( 5 x - 25 ) : ( 7 x - 25 ) = 5 y : 9 y 5 x - 25 = 5 y 7 x - 25 = 9 y 175 - 125 = 45 y - 35 y 50 = 10 y y = 5 5 x - 25 = 25 x = 10 the two numbers are 5 x and 7 x which are 50 and 70 . the answer is b ." | a = 9 * 25
b = 5 * 25
c = a - b
d = c * 5
e = 7 / 5
f = d * e
g = 9 * 25
h = 5 * 25
i = g - h
j = i * 5
k = f - j
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a ) s . 250 , b ) s . 280 , c ) s . 290 , d ) s . 260 , e ) s . 300 | d | divide(multiply(572, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 572 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 572 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 572 β y [ 12 / 10 + 1 ] = 572 β 22 y / 10 = 572 β 22 y = 5720 β y = 5720 / 22 = 520 / 2 = rs . 260 d" | a = 1 + 4
b = a * 2
c = 572 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 14 , b ) 16 , c ) 18 , d ) 12 , e ) 20 | c | subtract(divide(24, divide(3, 7)), 24) | a certain lab experiments with white and brown mice only . in one experiment , 3 / 7 of the mice are white . if therare 24 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 3 / 7 m number of brown mice = 4 / 3 m = 18 = > m = 18 answer c" | a = 3 / 7
b = 24 / a
c = b - 24
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a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 1 | b | divide(104, 42) | how many of the positive factors of 42 , 104 and how many common factors are there in numbers ? | "factors of 42 - 1,2 , 3,6 , 7,14 , 21,42 factors of 104 - 1,2 , 4,8 , 13,26 , 52,104 comparing both , we have two common factors of 40 and 104 - 1,2 , answer ( b )" | a = 104 / 42
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a ) 45 ( 4 / 11 ) % , b ) 45 % , c ) 53 ( 11 / 13 ) % , d ) 44 ( 5 / 11 ) % , e ) none of these | c | multiply(divide(subtract(130, add(multiply(3, const_4), multiply(8, 3))), 130), const_100) | a batsman scored 130 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : total runs scored = 130 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 130 - 60 = 70 required % = ( 70 / 130 ) Γ 100 = 700 / 13 = 53 ( 11 / 13 ) % answer : option c" | a = 3 * 4
b = 8 * 3
c = a + b
d = 130 - c
e = d / 130
f = e * 100
|
a ) $ 153 , b ) $ 698 , c ) $ 398 , d ) $ 549 , e ) $ 675 | b | subtract(815, divide(multiply(subtract(854, 815), 3), 4)) | a sum of money at simple interest amounts to $ 815 in 3 years and to $ 854 in 4 years . the sum is : | "b $ 698 s . i . for 1 year = $ ( 854 - 815 ) = $ 39 . s . i . for 3 years = $ ( 39 x 3 ) = $ 117 . principal = $ ( 815 - 117 ) = $ 698 ." | a = 854 - 815
b = a * 3
c = b / 4
d = 815 - c
|
a ) 14 , b ) 16 , c ) 10 , d ) 11 , e ) 12 | a | divide(434, subtract(32, const_1)) | along a yard 434 metres long , 32 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 32 trees have 31 gaps between them , required distance ( 434 / 31 ) = 14 option a" | a = 32 - 1
b = 434 / a
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a ) 20029 , b ) 20000 , c ) 20289 , d ) 20027 , e ) 15000 | e | divide(300, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 300 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 300 = > x = 15000 answer : e" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 300 / h
|
a ) 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35 | a | subtract(7, divide(add(multiply(7, const_2), 11), subtract(7, const_2))) | the product of a and b is equal to 11 more than twice the sum of a and b . if b = 7 , what is the value of b - a ? | "given : ab = 2 ( a + b ) + 11 and b = 7 hence 7 a = 2 a + 14 + 11 5 a = 25 , a = 5 ( a potential trap . do not get excited here . we need to find b - a , not a ) b - a = 7 - 5 = 2 correct option : a" | a = 7 * 2
b = a + 11
c = 7 - 2
d = b / c
e = 7 - d
|
a ) 60 , b ) 80 , c ) 50 , d ) 120 , e ) 140 | c | divide(add(5, 30), divide(subtract(const_100, 30), const_100)) | thirty percent of the members of a swim club have passed the lifesaving test . among the members who havenotpassed the test , 5 have taken the preparatory course and 30 have not taken the course . how many members are there in the swim club ? | "30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 30 + 5 = 42 members have not passed the test , thus 0.7 * total = 35 - - > total = 50 . answer : c ." | a = 5 + 30
b = 100 - 30
c = b / 100
d = a / c
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a ) 23 , b ) 22 , c ) 21 , d ) 24 , e ) 26 | e | divide(const_1, add(divide(const_1, 20), divide(divide(const_1, 20), 2))) | a is 2 times as fast as b . a alone can do the work in 20 days . if a and b work on alternative days , beginning with a , in how many days will the work be completed ? | "a can finish 1 work in 20 days b can finish 1 / 2 work in 20 days - since a is 2 faster than b this means b can finish 1 work in 20 * 2 days = 40 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 20 * 40 / ( 20 + 40 ) = 13 days since they alternate on days , it will take them double this time so total time = 13 * 2 = 26 days so answer is e" | a = 1 / 20
b = 1 / 20
c = b / 2
d = a + c
e = 1 / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | add(power(const_2, const_2), const_2) | how many different positive integers are factors of 125 ? | "125 = 5 * 25 answer : b" | a = 2 ** 2
b = a + 2
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | floor(divide(30, divide(10, const_2))) | what is the greatest integer m for which the number 30 ! / 10 ^ m is an integer ? | "10 ^ m = 2 ^ m * 5 ^ m . let ' s figure out how many 5 ' s are in the prime factorization of 30 ! the multiples of 5 are : 5 , 10 , 15 , 20 , 5 * 5 , 30 . thus 5 ^ 7 will divide 30 ! but 5 ^ 8 will not . clearly 2 ^ 7 will divide 30 ! so m = 7 is the largest possible integer . the answer is a ." | a = 10 / 2
b = 30 / a
c = math.floor(b)
|
a ) 10003 , b ) 10004 , c ) 10007 , d ) 10009 , e ) 10010 | b | add(multiply(const_1000, const_100), subtract(41, reminder(multiply(const_1000, const_100), 41))) | the smallest 5 digit number exactly divisible by 41 is ? | "the smallest 5 - digit number = 10000 . 41 ) 10000 ( 243 82 - - - 180 164 - - - - 160 123 - - - 37 - - - required number = 10000 + ( 41 - 37 ) = 10004 . option b" | a = 1000 * 100
b = 1000 * 100
c = 41 - reminder
d = a + c
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a ) 6 , b ) 5 , c ) 2 , d ) 8 , e ) 2 | d | divide(216, multiply(const_3, 9)) | 9 men are equal to as many women as are equal to 7 boys . all of them earn rs . 216 only . men β s wages are ? | "9 m = xw = 7 b 9 m + xw + 7 b - - - - - 216 rs . 9 m + 9 m + 9 m - - - - - 216 rs . 27 m - - - - - - 216 rs . = > 1 m = 8 rs . answer : d" | a = 3 * 9
b = 216 / a
|
a ) 10.51 , b ) 11.51 , c ) 9.51 , d ) 13.51 , e ) 15.51 | a | divide(80, multiply(add(20, 4), const_0_2778)) | the speed at which a man can row a boat in still water is 20 kmph . if he rows downstream , where the speed of current is 4 kmph , what time will he take to cover 80 metres ? | speed of the boat downstream = 20 + 4 = 24 kmph = 24 * 5 / 18 = 6.66 m / s hence time taken to cover 70 m = 70 / 6.66 = 10.51 seconds . answer : a | a = 20 + 4
b = a * const_0_2778
c = 80 / b
|
a ) 180 , b ) 882 , c ) 277 , d ) 261 , e ) 281 | a | multiply(360, divide(360, add(add(5, 12), 13))) | the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 360 m , its area is ? | "5 x + 12 x + 13 x = 360 = > x = 12 a = 60 , b = 144 , c = 156 s = ( 60 + 144 + 156 ) / 2 = 180 answer : a" | a = 5 + 12
b = a + 13
c = 360 / b
d = 360 * c
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a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 600 | e | subtract(divide(subtract(multiply(3, 1800), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1800), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | the cost of 2 chairs and 3 tables is rs . 1800 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "explanation : 2 c + 3 t = 1800 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 600 = > t - c = 600 answer : e" | a = 3 * 1800
b = 2 * 3
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
h = 3 * 1800
i = 2 * 3
j = h - i
k = 3 * 3
l = 2 * 2
m = k - l
n = j / m
o = 2 * n
p = 3 - o
q = p / 3
r = g - q
|
a ) $ 496.30 , b ) $ 512.40 , c ) $ 555.40 , d ) $ 574.90 , e ) $ 588.20 | a | divide(670, add(divide(35, const_100), const_1)) | jim is able to sell a hand - carved statue for $ 670 which was a 35 % profit over his cost . how much did the statue originally cost him ? | "670 is the final price that is the selling price . now suppose , x is the cost price . therefore 670 = 1.35 * x now while doing x = 670 / 1.35 calculation . i saw that the answer choices begin with 5 . so i did 135 * 5 = 675 . as 670 is less than 675 . therefore the quotient will be less than 5 . which is only in the answer a ." | a = 35 / 100
b = a + 1
c = 670 / b
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a ) 5 : 1 , b ) 1 : 1 , c ) 20 : 4 , d ) 10 : 3 , e ) 2 : 1 | e | divide(10, 5) | a baseball team played 10 games and won 5 . what is the ratio of the number of games played to the number of losses ? | the number of games played is 10 . the number of games won is 5 , so 10 - 5 = 5 = number of game losses . answer = ! 0 : 5 = 2 : 1 answer is e | a = 10 / 5
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a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | b | divide(multiply(48, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 15), multiply(divide(add(24, 40), 4), 3)), multiply(divide(8, 2), 3))) | 3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 48 days . if 40 more cows and 15 more buffaloes are brought in , how many days will the fodder last ? | "2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 48 days . 40 cows + 15 buffaloes = 20 oxen + 10 oxen = 30 oxen the new total is equal to 60 oxen instead of 30 oxen . 30 * 48 / 60 oxen = 24 days the answer is b ." | a = 24 / 4
b = a * 3
c = 15 + b
d = 8 / 2
e = d * 3
f = c + e
g = 48 * f
h = 15 + 15
i = 24 + 40
j = i / 4
k = j * 3
l = h + k
m = 8 / 2
n = m * 3
o = l + n
p = g / o
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a ) $ 10,500 , b ) $ 10,500 , c ) $ 11,500 , d ) $ 12,000 , e ) $ 12,500 | e | subtract(multiply(multiply(2, const_100), sqrt(const_100)), divide(multiply(multiply(2, const_100), sqrt(const_100)), add(divide(add(divide(19, const_100), 1), subtract(1, divide(15, const_100))), 1))) | cp of 2 buffaloes are $ 30,000 . by selling 1 at a loss of 15 % and other at a profit of 19 % , he foundthat sp of both buffaloes is the same . find the cp of each ? | "let the c . p of 1 st buffalo = x β΄ c . p of 2 nd buffalo = 30,000 - x s . p of 1 st buffalo = ( 100 - loss % ) - - - - - - - - - - - - - - - - - - - x c . p 100 s . p = 100 - 15 x - - - - - - - - - - - = 85 x / 100 100 s . p of 2 nd buffalo = ( 100 - profit % ) - - - - - - - - - - - - - - - - - - - - - x c . p 100 s . p = ( 100 + 19 ) ( 30,000 - x ) 119 ( 30,000 - x ) - - - - - - - - - - - - - - - - - - - - = - - - - - - - - - - - - - - - - 100 100 s . p of 1 st buffalo = s . p of 2 nd buffalo 85 x / 100 = 119 ( 30,000 - x ) / 100 β 85 x = 119 ( 30,000 - x ) β 85 x = 3 , 570,000 - 119 x β 85 x + 119 x = 3 , 570,000 β 204 x = 3 , 570,000 β x = 3 , 570,000 / 204 β x = $ 17,500 c . p of 1 st buffalo = $ 17,500 c . p of 2 nd buffalo = 30,000 - 17,500 = $ 12,500 e" | a = 2 * 100
b = math.sqrt(100)
c = a * b
d = 2 * 100
e = math.sqrt(100)
f = d * e
g = 19 / 100
h = g + 1
i = 15 / 100
j = 1 - i
k = h / j
l = k + 1
m = f / l
n = c - m
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a ) 15 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | c | add(multiply(divide(12, 48), const_100), 10) | the purchase price of an article is $ 48 . in order to include 10 % of cost for overhead and to provide $ 12 of net profit , the markup should be | "cost price of article = 48 $ % of overhead cost = 10 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 10 = 35 % answer c" | a = 12 / 48
b = a * 100
c = b + 10
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a ) 134 , b ) 146 , c ) 152 , d ) 162 , e ) 170 | a | multiply(multiply(multiply(200, subtract(1, divide(1, 8))), subtract(1, divide(1, 8))), subtract(1, divide(1, 8))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 8 . for a group of 200 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "the probability of survival for each of the first 3 months of life is 1 - 1 / 8 = 7 / 8 , so of 200 newborn 200 * 7 / 8 * 7 / 8 * 7 / 8 = ~ 134 is expected to survive . answer : a ." | a = 1 / 8
b = 1 - a
c = 200 * b
d = 1 / 8
e = 1 - d
f = c * e
g = 1 / 8
h = 1 - g
i = f * h
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a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 43.75 | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(divide(15, const_100), const_1)), const_1), const_100) | increasing the original price of a certain item by 15 percent and then increasing the new price by 25 percent is equivalent to increasing the original price by what percent ? | "we ' re told that the original price of an item is increased by 15 % and then that price is increased by 25 % . . . . if . . . . starting value = $ 100 + 15 % = 100 + . 15 ( 100 ) = 115 + 25 % = 115 + . 25 ( 115 ) = 115 + 28.75 = 143.75 the question asks how the final price relates to the original price . this is essentially about percentage change , which means we should use the percentage change formula : percentage change = ( new - old ) / old = difference / original doing either calculation will yield the same result : 43.75 / 100 = 43.75 % final answer : e" | a = 25 / 100
b = a + 1
c = 15 / 100
d = c + 1
e = b * d
f = e - 1
g = f * 100
|
a ) 4 , b ) 5 , c ) 7 , d ) 6 , e ) 8 | b | subtract(378461, multiply(floor(divide(378461, 13)), 13)) | find the least number must be subtracted from 378461 so that remaining no . is divisible by 13 ? | "on dividing 378461 by 13 we get the remainder 5 , so 5 should be subtracted b" | a = 378461 / 13
b = math.floor(a)
c = b * 13
d = 378461 - c
|
a ) $ 480 , b ) $ 450 , c ) $ 896 , d ) $ 870 , e ) $ 660 | c | add(divide(multiply(divide(add(832, 448), const_2), 40), const_100), divide(add(832, 448), const_2)) | the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 40 % profit ? | let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 140 % of $ 640 = $ 896 . c | a = 832 + 448
b = a / 2
c = b * 40
d = c / 100
e = 832 + 448
f = e / 2
g = d + f
|
a ) 20.8 , b ) 21.46 , c ) 21.74 , d ) 22.3 , e ) 21.4 | c | multiply(const_100, divide(subtract(28.00, multiply(11.50, 2)), multiply(11.50, 2))) | a certain pair of used shoes can be repaired for $ 11.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 28.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? | "1 ) cost of repairing = 11.5 ( for one year ) , therefore for 2 years it would be $ 23 . 2 ) cost of new pair which will last for 2 years is $ 28 . percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoeinitial value would be the cost of repairing the old shoe . i . e ( 28 - 23 ) / ( 23 ) * 100 = 21.74 % . ans is c" | a = 11 * 50
b = 28 - 0
c = 11 * 50
d = b / c
e = 100 * d
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a ) 16 , b ) 18 , c ) 40 , d ) 24 , e ) 30 | c | multiply(const_3.0, divide(80, add(4, 6))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 80 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 80 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 80 / 10 = 8 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 8 = 40 kms . . . answer - c" | a = 4 + 6
b = 80 / a
c = 3 * 0
|
a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 17 | b | inverse(subtract(divide(const_1, 4), divide(const_1, 6))) | renu can do a piece of work in 6 days , but with the help of her friend suma , she can do it in 4 days . in what time suma can do it alone ? | "renu β s one day β s work = 1 / 6 suma β s one day β s work = 1 / 4 - 1 / 6 = 1 / 12 suma can do it alone in 12 days . answer : b" | a = 1 / 4
b = 1 / 6
c = a - b
d = 1/(c)
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a ) 250 , b ) 425 , c ) 675 , d ) 825 , e ) 750 | d | multiply(add(divide(100, 8), divide(150, 10)), 30) | working at their respective constant rates , machine a makes 100 copies in 8 minutes and machine b makes 150 copies in 10 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ? | "machine a can produce 100 * 30 / 8 = 375 copies and , machine b can produce 150 * 30 / 10 = 825 copies total producing 700 copies . d is the answer" | a = 100 / 8
b = 150 / 10
c = a + b
d = c * 30
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a ) 25 , b ) 36 , c ) 38 , d ) 45 , e ) 42 | e | add(divide(subtract(multiply(subtract(84, 25), 1.5), 11), add(1.5, const_1)), 11) | q - 1 ) of the 84 parents who attended a meeting at a school , 25 volunteered to supervise children during the school picnic and 11 volunteered both to supervise children during the picnic and to bring refreshments to the picnic . if the number of parents who volunteered to bring refreshments was 1.5 times the number of parents who neither volunteered to supervise children during the picnic nor volunteered to bring refreshments , how many of the parents volunteered to bring refreshments ? | i used the following formula : total = group 1 + group 2 + neither - both using the information from the question stem , we have : 84 total people 25 who supervise ( group 1 ) x who neither supervise nor bring refreshments 1.5 x people who bring refreshments ( group 2 ) 11 who supervise and bring refreshments ( both ) therefore . . . 84 = 25 + 1.5 x + x - 11 solving , we get x = 28 since we want the value of those who bring refreshments and supervise : 1.5 ( 28 ) = 42 answer is e | a = 84 - 25
b = a * 1
c = b - 11
d = 1 + 5
e = c / d
f = e + 11
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a ) 18000 , b ) 22000 , c ) 24000 , d ) 25000 , e ) 32000 | d | multiply(multiply(subtract(1, divide(4, 6)), multiply(multiply(multiply(const_0_25, const_100), const_100), const_10)), add(divide(1, 4), 1)) | in a mayoral election , candidate x received 1 / 4 more votes than candidate y , and candidate y received 1 / 6 fewer votes than z . if z received 24,000 votes how many votes did candidate x received ? | "z = 24 - - > y received 1 / 4 fewer votes than z - - > y = z - 1 / 6 * z = 20 ; x received 1 / 4 more votes than y - - > x = y + 1 / 4 * y = 25 . answer : d ." | a = 4 / 6
b = 1 - a
c = const_0_25 * 100
d = c * 100
e = d * 10
f = b * e
g = 1 / 4
h = g + 1
i = f * h
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a ) 7 min , b ) 8 min , c ) 9 min , d ) 10 min , e ) 55.8 min | e | multiply(const_60, divide(subtract(72, 45), 72)) | excluding stoppages , the speed of a bus is 72 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 9 km less . time taken to cover 67 km = ( 67 / 72 ) x 60 = 55.8 min answer : e" | a = 72 - 45
b = a / 72
c = const_60 * b
|
a ) 4 , b ) 4.5 , c ) 5 , d ) 5.5 , e ) 6 | d | divide(add(multiply(60, const_4), multiply(divide(multiply(const_4, 60), add(const_1, const_4)), const_4)), add(60, divide(multiply(const_4, 60), add(const_1, const_4)))) | two cars are driving toward each other . the first car is traveling at a speed of 60 km / h , which is 25 % slower than the second car ' s speed . if the distance between the cars is 770 km , how many hours will it take until the two cars meet ? | "the speed of the first car is 60 km / h . the speed of the second car is 60 / 0.75 = 80 km / h . the two cars complete a total of 140 km each hour . the time it takes the cars to meet is 770 / 140 = 5.5 hours . the answer is d ." | a = 60 * 4
b = 4 * 60
c = 1 + 4
d = b / c
e = d * 4
f = a + e
g = 4 * 60
h = 1 + 4
i = g / h
j = 60 + i
k = f / j
|
a ) 65 , b ) 87 , c ) 71 , d ) 60 , e ) 66 | c | subtract(107, multiply(4, subtract(15, 6))) | in 15 year β s time the combined age of my 4 brothers will be 107 . what will it be in 6 year β s time ? | c 71 combined age in 15 years = 107 . 4 Γ 15 = 60 , therefore combined age now is 107 β 60 = 47 . in 6 years time , therefore , combined age will be 47 + 24 ( 4 Γ 6 ) = 71 . | a = 15 - 6
b = 4 * a
c = 107 - b
|
a ) rs . 424 , b ) rs . 524 , c ) rs . 324 , d ) rs . 350 , e ) rs . 354 | c | add(divide(330, add(const_1, divide(10, const_100))), multiply(divide(8, const_100), divide(330, add(const_1, divide(10, const_100))))) | if sharon ' s weekly salary increased by 10 percent , she would earn $ 330 per week . if instead , her weekly salary were to increase by 8 percent , how much would she earn per week ? | "soln : - ( 330 / 110 ) 108 = 324 in this case long division does not take much time . ( 330 / 110 ) 108 = 324 answer : c" | a = 10 / 100
b = 1 + a
c = 330 / b
d = 8 / 100
e = 10 / 100
f = 1 + e
g = 330 / f
h = d * g
i = c + h
|
a ) 2.5 , b ) 3 1 / 4 , c ) 3 , d ) 2 3 / 4 , e ) 4 | b | divide(27,000, 11,000) | the compound interest on $ 27,000 at 11 % p . a is $ 11,000 . find out the period ( in years ) ? | "amount = 27,000 + 11,000 = 38,000 . . . . . . . . . now by using compound interest formula we get , 27,000 * ( 111 / 100 ) ^ n = 38000 = = = = > n = = 3 1 / 4 years ans - b" | a = 27 / 0
|
a ) 26 , b ) 30 , c ) 31 , d ) 41 , e ) 43 | d | multiply(divide(1,5, 13,25), const_100) | 1,5 , 13,25 , __ | "series increased by 1 1 + ( 4 * 1 ) = 1 + 4 = 5 5 + ( 4 * 2 ) = 5 + 8 = 13 13 + ( 4 * 3 ) = 13 + 12 = 25 25 + ( 4 * 4 ) = 25 + 16 = 41 answer : d" | a = 1 / 5
b = a * 100
|
a ) 6,24 , b ) 9,27 , c ) 10,28 , d ) 18,36 , e ) 19,37 | c | subtract(divide(subtract(add(multiply(4, 4), 18), 4), subtract(4, const_1)), const_1) | ages of two persons differ by 18 years . if 4 years ago , the elder one was 4 times as old the younger one , find their present age | explanation : let the age of younger person be x , then elder person ' s age is ( x + 18 ) = > 4 ( x - 4 ) = ( x + 18 - 4 ) [ 4 years before ] = > 4 x - 16 = x + 14 = > x = 10 . so elder person ' s age is x + 18 = 28 answer : option c | a = 4 * 4
b = a + 18
c = b - 4
d = 4 - 1
e = c / d
f = e - 1
|
a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 4 | e | multiply(divide(13, add(13, 13)), 4) | if 13 = 13 w / ( 1 - w ) , then ( 4 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 4 w = 2 4 w * 2 = 2 * 2 = 4 answer : e" | a = 13 + 13
b = 13 / a
c = b * 4
|
a ) 2,000 , b ) 3,000 , c ) 6,000 , d ) 8,000 , e ) 9,000 | a | multiply(divide(divide(subtract(subtract(multiply(multiply(2, const_4), const_1000), multiply(multiply(multiply(2, const_4), const_1000), divide(2, 3))), multiply(subtract(multiply(multiply(2, const_4), const_1000), multiply(multiply(multiply(2, const_4), const_1000), divide(2, 3))), divide(2, 5))), const_1000), const_4), 2) | in a recent head - to - head run - off election , 10,000 absentee ballets were cast . 2 / 3 of the absentee ballets were thrown out and 2 / 5 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | "1 / 3 * 3 / 5 ( total absentee votes ) = 1 / 5 ( total votes ) = 1 / 5 * 10000 = 2000 answer is a" | a = 2 * 4
b = a * 1000
c = 2 * 4
d = c * 1000
e = 2 / 3
f = d * e
g = b - f
h = 2 * 4
i = h * 1000
j = 2 * 4
k = j * 1000
l = 2 / 3
m = k * l
n = i - m
o = 2 / 5
p = n * o
q = g - p
r = q / 1000
s = r / 4
t = s * 2
|
a ) 246 , b ) 522 , c ) 1408 , d ) 710 , e ) 223 | b | divide(multiply(const_1000, 1), multiply(add(const_3, divide(add(multiply(const_3, const_4), const_2), power(add(multiply(const_4, const_2), const_2), const_2))), 0.61)) | a bicycle wheel has a diameter of 0.61 m . how many complete revolutions does it make in 1 km ? | "1 revolution = 3.14 * diameter . number of revolutions in 1 km = 1000 m / ( 3.14 * 0.61 m ) = 522.1 . hence 522 complete revolutions . answer b" | a = 1000 * 1
b = 3 * 4
c = b + 2
d = 4 * 2
e = d + 2
f = e ** 2
g = c / f
h = 3 + g
i = h * 0
j = a / i
|
a ) 8000 , b ) 2787 , c ) 27766 , d ) 9976 , e ) 21671 | a | divide(multiply(8160, const_100), add(add(multiply(6, 3), multiply(9, 5)), multiply(13, 3))) | nitin borrowed some money at the rate of 6 % p . a . for the first 3 years , 9 % p . a . for the next 5 years and 13 % p . a . for the period beyond 8 years . if the total interest paid by him at the end of 11 years is rs . 8160 , how much money did he borrow ? | let the sum be rs . x . then , [ ( x * 6 * 3 ) / 100 ] + [ ( x * 9 * 5 ) / 100 ] + [ ( x * 13 * 3 ) / 100 ] = 8160 18 x + 45 x + 39 x = ( 8160 * 100 ) 102 x = 816000 = > x = 8000 . answer : a | a = 8160 * 100
b = 6 * 3
c = 9 * 5
d = b + c
e = 13 * 3
f = d + e
g = a / f
|
a ) 23 , b ) 1 : 2 , c ) 2 : 5 , d ) 3 : 5 , e ) 2 : 1 | e | divide(subtract(6, 2), 2) | the ratio between the sale price and the cost price of an article is 6 : 2 . what is the ratio between the profit and the cost price of that article ? | "c . p . = rs . 2 x and s . p . = rs . 6 x . then , gain = rs . 4 x required ratio = 4 x : 2 x = 2 : 1 e" | a = 6 - 2
b = a / 2
|
a ) 60 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | add(40, divide(subtract(1200, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1200 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 1200 - 640 = 560 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 560 / 28 = 20 total hrs = 40 + 20 = 60 answer a 60" | a = 16 * 40
b = 1200 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 11 , b ) 21 , c ) 22 , d ) 29 , e ) 32 | d | subtract(multiply(const_2, 15), const_1) | if 5 ^ 29 * 4 ^ 15 = 2 * 10 ^ n , what is the value of n ? | my attempt : 5 ^ ( 29 ) * 4 ^ ( 15 ) = 2 * 10 ^ p 5 ^ ( 29 ) * 2 ^ ( 30 ) = 2 * ( 2 * 5 ) ^ p 5 ^ ( 29 ) * 2 ^ ( 30 ) = 2 * 2 ^ ( p ) * 5 ^ ( p ) 5 ^ ( 29 ) * 2 ^ ( 30 ) = 2 ^ ( p + 1 ) * 5 ^ ( p ) = > ( p + 1 ) = 30 = > p = 29 answer : 29 ( d ) | a = 2 * 15
b = a - 1
|
a ) 29 , b ) 45 , c ) 81 , d ) 64 , e ) 18 | d | multiply(power(4, 2), power(2, 2)) | if 2 ^ k = 4 , then 2 ^ ( 2 k + 2 ) = | "2 ^ k = 4 2 ^ 2 k = 4 ^ 2 2 ^ 2 k = 16 2 ^ ( 2 k + 2 ) = 2 ^ 2 k * 2 ^ 2 = 16 * 4 = 64 answer : d" | a = 4 ** 2
b = 2 ** 2
c = a * b
|
a ) 13 , b ) 33 , c ) 50 , d ) 51 , e ) 60 | e | multiply(multiply(60, divide(1, add(1, 2))), add(1, 2)) | a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 60 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ? | "yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 60 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 30 cents . fraction of spirit is 1 / 2 . 30 = k * ( 1 / 2 ) k = 60 cost per liter = 60 * ( 1 / 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 60 * ( 1 / 3 ) * 3 = 50 cents e . 60 cents" | a = 1 + 2
b = 1 / a
c = 60 * b
d = 1 + 2
e = c * d
|
a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 52.1 | c | divide(add(multiply(28, 40), multiply(50, 60)), add(28, 50)) | the average marks of a class of 28 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 28 students = 28 * 40 = 1120 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 78 students = 1120 + 3000 = 4120 average marks of all the students = 4120 / 78 = 52.8 answer : c" | a = 28 * 40
b = 50 * 60
c = a + b
d = 28 + 50
e = c / d
|
a ) 256 , b ) 512 , c ) 432 , d ) 352 , e ) 640 | b | multiply(multiply(divide(16, const_2), divide(16, const_2)), divide(16, const_2)) | free notebooks were distributed equally among children of a class . the number of notebooks each child got was one - eighth of the number of children . had the number of children been half , each child would have got 16 notebooks . total how many notebooks were distributed ? | let total number of children be x . then , x * ( 1 / 8 ) x = ( x / 2 ) x 16 x = 64 . number of notebooks = ( 1 / 8 ) x ^ 2 = ( 1 / 8 ) x 64 x 64 = 512 . answer is b . | a = 16 / 2
b = 16 / 2
c = a * b
d = 16 / 2
e = c * d
|
a ) 21 , b ) 28 , c ) 26 , d ) 25 , e ) 24 | e | divide(multiply(2400, 3.25), 7800) | if 7800 / 3.25 = 2400 , then 780.0 / 32.5 is equal to ? | "answer given expression 780.0 / 32.5 = 7800 / 325 = 7800 / ( 3.25 x 100 ) = 2400 / 100 = 24 correct option : e" | a = 2400 * 3
b = a / 7800
|
a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 60 miles . , e ) 80 miles . | e | multiply(divide(10, add(5, 10)), 120) | tim and Γ© lan are 120 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and Γ© lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets Γ© lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice e meet the 2 : 1 ( tim : elan = 80 : 40 ) ratio of maintaining total distance traveled 120 miles so correct answer e" | a = 5 + 10
b = 10 / a
c = b * 120
|
a ) 6,500 , 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 40 positive even integers is 2,550 . 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 = 40 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) sum of 1 st 100 natural numbers = ( 100 * 101 ) / 2 = 5050 sum of 1 st 40 positive even integers = 2550 sum of 1 st 100 odd numbers = 5050 - 2550 = 2500 so 101 + 103 + . . . . . . . 199 = 40 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) = 4000 + 2500 = 6500 a is the answer ." | a = 200 + 101
b = a / 2
c = 200 - 101
d = c / 2
e = d + 1
f = b * e
|
a ) 14 , b ) 28 , c ) 63 , d ) 84 , e ) 100 | e | add(10, const_1) | if x and y are positive integers and 20 x = 10 y what is the least possible value of xy ? | "20 x = 10 y = > x / y = 1 / 2 = > 2 x = y 2 ( 3 ) = 1 ( 3 ) = > x * y = 9 but it is not given 2 ( 10 ) = 1 ( 10 ) = > x * y = 100 e" | a = 10 + 1
|
a ) 125 , b ) 127 , c ) 129 , d ) 131 , e ) 150 | b | gcd(subtract(1661, 10), subtract(2045, 13)) | find the greatest number which on dividing 1661 and 2045 , leaves a reminder of 10 and 13 respectively | explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 1661 - 10 ) and ( 2045 - 13 ) = hcf ( 1651 , 2032 ) = 127 option b | a = 1661 - 10
b = 2045 - 13
c = math.gcd(a, b)
|
a ) 22678 , b ) 26699 , c ) 26788 , d ) 19000 , e ) 39000 | e | subtract(40000, 1000) | the price of a t . v . set worth rs . 40000 is to be paid in 20 installments of rs . 1000 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 . 1000 balance payment = ( 40000 - 1000 ) = rs . 39000 answer : e" | a = 40000 - 1000
|
a ) 45 % , b ) 49 % , c ) 54 % , d ) 59 % , e ) 61 % | c | multiply(divide(11628, add(add(2136, 7636), 11628)), const_100) | 3 candidates in an election and received 2136 , 7636 and 11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 2136 + 7636 + 11628 ) = 21260 so , required percentage = 11628 / 21260 * 100 = 54 % c" | a = 2136 + 7636
b = a + 11628
c = 11628 / b
d = c * 100
|
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