The Shopkeeper increased the price of a product by 75% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 140% of the required amount. What is the net difference in the expenditure on that product? a) 12.5 b) 26.0 c) 13.5 d) 17.5 e) 19.8

The Shopkeeper increased the price of a product by 75% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 140% of the required amount. What is the net difference in the expenditure on that product? a) 12.5 b) 26.0 c) 13.5 d) 17.5 e) 19.8
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1 Answer

Answer :

Answer: D

 Quantity×Rate=Price

 1x1=1

 1.4x1.75=2.45

 Decrease in price = (0.175/1) × 100

 = 17.5%
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Related questions

Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4

Description : Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4 

Last Answer : Answer: D  Let the marks secured by them be x and (x + 18)  Then sum of their marks = x + (x + 18) = 2x + 18  Given that (x + 18) was 72% of the sum of their marks  =>(x+18) = 72/100(2x+18)  => ... 11x = 126 x = 11.45  Then (x + 18) = 11.45 + 18 = 29.45  Hence their marks are 11.45 and 29.45

The value of a machine depreciates at the rate of 20% every year. It was purchased 6 years ago. If its present value is Rs. 17,496, its purchase price was : a) 25023.4 b) 67040.0 c) 34171.8 d) 27337.5

Description : The value of a machine depreciates at the rate of 20% every year. It was purchased 6 years ago. If its present value is Rs. 17,496, its purchase price was : a) 25023.4 b) 67040.0 c) 34171.8 d) 27337.5

Last Answer : Answer: C  Purchase price = 17,496 / (1 -20/ 100)³  = 17,496 × 10/8 × 10/8 ×10/8  = 17496 × 1.25 × 1.25 ×1.25  = 17496 ×1.953  = 34171.8

Chenna dhal is now being sold at Rs. 70 a kg. Last month, is rate was Rs. 80 per kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed? a)12.5 b)21.8 c)23 d)18

Description : Chenna dhal is now being sold at Rs. 70 a kg. Last month, is rate was Rs. 80 per kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed? a)12.5 b)21.8 c)23 d)18

Last Answer : Answer: A  Let a family's monthly consumption of chenna dhal be x kg.  To keep the expenditure fixed,  their consumption for this month should be 70x/80 = 7x/8.  Reduction in consumption = x/8 = 12.5% of x

Harshika wants to purchase a mobile handset. The shopkeepers told her to pay 30% tax if she asked for the actual sale price (without tax) of the mobile and paid to the shopkeeper Rs. 16150. In the doing so, she managed to avoid to pay the 30% tax. What is the amount of discount that she received on the selling price (inclusive of tax)?

Description : Harshika wants to purchase a mobile handset. The shopkeepers told her to pay 30% tax if she asked for the actual sale price (without tax) of the mobile and paid to the shopkeeper Rs. 16150. In the doing so, she managed to avoid ... tax)? 1. Rs. 5495 2. Rs. 5850 3. Rs. 5950 4. Rs. 5685 5. Rs. 5975

Last Answer : 3; SP = 100, SP (with tax) = 130 New SP = 100 - 5 = 95 Effective discount = 130 - 95 = 35 So, when SP of 95, discount = 35 And when SP of Rs. 16,150, discount = 35/95x16150 = Rs. 5950

A shopkeeper bought 1800 blackberry and 1200 blueberry. He found 45% of blackberry and 24% of blueberry were rotten. Find the percentage of fruits in good condition. a) 63.4 b) 32.9 c) 48.5 d) 56.3

Description : A shopkeeper bought 1800 blackberry and 1200 blueberry. He found 45% of blackberry and 24% of blueberry were rotten. Find the percentage of fruits in good condition. a) 63.4 b) 32.9 c) 48.5 d) 56.3

Last Answer :  Answer: A Total number of fruits shopkeeper bought = 1800 + 1200 = 1000 Number of rotten blackberry = 45% of 1800  = 45/100 1800  = 81000/100  = 810 Number of rotten blueberry = 24% of 1200 ... Therefore Percentage of fruits in good condition = (1902/3000 100)  = (190200/3000)  = 63.4%

An shopkeeper has 15 models of cup and 9 models of saucer. In how many ways can he make a pair of cup and saucer? A) 100 B) 80 C) 110 D) 135

Description : An shopkeeper has 15 models of cup and 9 models of saucer. In how many ways can he make a pair of cup and saucer? A) 100 B) 80 C) 110 D) 135

Last Answer : Answer: D) He has 15 patterns of cup and 9 model of saucer A cup can be selected in 15 ways. A saucer can be selected in 9 ways. Hence one cup and one saucer can be selected in 15×9 ways =135 ways

A number is increased by 70 % and then decreased by 70 %. Find the net increase or decrease per cent. a) 25 b) 56 c) 45 c) 49 e) 98

Description : A number is increased by 70 % and then decreased by 70 %. Find the net increase or decrease per cent. a) 25 b) 56 c) 45 c) 49 e) 98 

Last Answer : Answer: D Let the number be 100. Increase in the number = 70 % = 40 % of 100  = (40/100 100)  = 70 Therefore, increased number = 100 + 70 = 170 This number is decreased by 70 % Therefore, decrease ... = 100 - 51 = 49 Hence, net percentage decrease = (49/100 100)  = (4900/100)  = 49 %

Mithun went to a shop and bought things worth Rs. 75, out of which 90 Paise went on sales tax on taxable purchases. If the tax rate was 18%, then what was the cost of the tax free items? a) 12.5 b) 69.80 c) 19.7 d) 34.5 e) 69.1

Description : Mithun went to a shop and bought things worth Rs. 75, out of which 90 Paise went on sales tax on taxable purchases. If the tax rate was 18%, then what was the cost of the tax free items? a) 12.5 b) 69.80 c) 19.7 d) 34.5 e) 69.1

Last Answer : Answer: E  Total cost of the items he purchased = Rs.75  Given that out of this Rs.75, 90 Paise is given as tax  => Total tax incurred = 90 Paise  = Rs.90/100  Let the cost of the tax free items = x  Given that tax rate = ... −0.9 −x) = 90  ⇒ (75 − 0.9 − x) = 5  x = 75 − 0.9 - 5  X=69.1 

The price of dhal is increased from $ 30 to $ 37.5 per kg. Find the percentage increase in price. a) 25 b) 60 c) 45 d) None e) 98

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Last Answer : Answer: A Price of dhal before = $30 Price of dhal now = $ 37.5 Increase in dhal = current price - original price  = $37.5 - $ 30  = $ 7.5 Therefore, percentage increase in price = Increase in price/ ...  = 7.5 / 30 100 = 750/100 = 25 % Thus, increase in price= 25 %

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

Gowthaman needs 25% to pass. If he scored 424 marks and falls short by 26 marks, what was the maximum marks he could have got? a) 2725 b) 2650 c) 1800 d) 1750 e) 989

Description : Gowthaman needs 25% to pass. If he scored 424 marks and falls short by 26 marks, what was the maximum marks he could have got? a) 2725 b) 2650 c) 1800 d) 1750 e) 989 

Last Answer : Answer: C  If Gowthaman had scored 26 marks more, he could have scored 25%  Therefore, Mike required 424 + 26 = 450 marks  Let the maximum marks be x.  Then 25 % of x = 450  (25/100) × x = 450  x = (450 × 100)/25 x = 45000/25 x = 1800

Vasavi spends 50% of her monthly income on grocery, clothes and education in the ratio of 8 : 4 : 10 respectively. If the amount spent on clothes is 2770/–, what is Vasavi's monthly income? a) 15235 b) 65000 c) 55400 d) 30470 e) 98700

Description : Vasavi spends 50% of her monthly income on grocery, clothes and education in the ratio of 8 : 4 : 10 respectively. If the amount spent on clothes is 2770/–, what is Vasavi's monthly income? a) 15235 b) 65000 c) 55400 d) 30470 e) 98700

Last Answer : Answer: D Ratio of Expenses = 8: 4: 10 therefore amount spend on clothes, i.e. 4x = 2770  x = 692.5  Total exp = (8 + 4 + 10)x  = 22x. = 22 × 692.5 = 15235  Monthly income be x.  50% of x = 15235  x = 15235 × 100/ 50  X = 30470 

A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Description : A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Last Answer : Katie earned 84,92,84,75 and 70 on her first 5 tests. What is the minimum grade Katie needs to earn on the next test to have a mean of 84?

A noodles merchant buys two varieties of noodles the price of the first being twice that of the second. He sells the mixture at Rs 17.50 per kilogram thereby making a profit of 25% . If the ratio of the amounts of the first noodles and the second noodles in the mixture is 2:3, then the respective costs of each noodles are A) Rs 20, Rs 10 B) Rs 24, Rs 12 C) Rs 16, Rs 8 D) Rs 26, Rs 13 E) NONE

Description : A noodles merchant buys two varieties of noodles the price of the first being twice that of the second. He sells the mixture at Rs 17.50 per kilogram thereby making a profit of 25% . If the ratio of the amounts of the first noodles ... , Rs 10 B) Rs 24, Rs 12 C) Rs 16, Rs 8 D) Rs 26, Rs 13 E) NONE

Last Answer : Answer: A

The population of a town was 200000 three years ago. If increased by 4%, 6% and 10% respectively in the last three years, then the present population of town is a) 1,10,313 b) 1,10,314 c) 2,42,528 d) 2,93,313 e) 2,10,313

Description : The population of a town was 200000 three years ago. If increased by 4%, 6% and 10% respectively in the last three years, then the present population of town is a) 1,10,313 b) 1,10,314 c) 2,42,528 d) 2,93,313 e) 2,10,313

Last Answer : Answer: C The present population = (1+4/100)×(1+6/100)×(1+10/100)×200000  = (104/100) × (106/100) × (110/100) × 200000  = 1.04×1.06×1.1×200000  = 2,42,528

Find the increase value if 450 is increased by 90 %. a) 525 b) 715 c) 645 d) 795 e) 855

Description : Find the increase value if 450 is increased by 90 %. a) 525 b) 715 c) 645 d) 795 e) 855

Last Answer : Answer: E  Increase = 90 % of 450  = 90/100 × 450  = 40500/100  = 405 Therefore, increase value = 450 + 405 = 855

Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41

Description : Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41 

Last Answer :  Answer: A (i) Percentage scored in Mathematics = 120/180 100  = 12000/180  =1200/18  = 66 2 /3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the ... percentage on the whole = (400/510 100)  = (40000/510)  = 4000/51  = 78 22/51 %

In a birthday party, every person shakes hand with every other person. If there was a total of 66 handshakes in the party, how many persons were present in the party? A) 9 B) 8 C) 7 D) 12

Description :  In a birthday party, every person shakes hand with every other person. If there was a total of 66 handshakes in the party, how many persons were present in the party? A) 9 B) 8 C) 7 D) 12

Last Answer : Answer: D)  Assume that in a party every person shakes hand with every other person  Number of hand shake = 66  Total number of hand shake is given by nC2  Let n = the total number of persons ... cannot take negative value of n  So, n = 12  Therefore number of persons in the party = 12

A question paper has two parts A and B, each containing 12 questions. If a student needs to choose 10 from part A and 8 from part B, in how many ways can he do that? A) 32670 B) 36020 C) 41200 D) 29450

Description : A question paper has two parts A and B, each containing 12 questions. If a student needs to choose 10 from part A and 8 from part B, in how many ways can he do that? A) 32670 B) 36020 C) 41200 D) 29450

Last Answer : Answer: A)  Number of ways to choose 10 questions from part A = 12C10  Number of ways to choose 8 questions from part B = 12C8  Total number of ways= 12C10 × 12C8  = 12C2 × 12C4 [∵ nCr = nC(n-r)]  ={12×11}/{2×1}X{12×11×10×9} / {4×3×2×1}  =66×495 =32670

Pramoth has 12 friends and he wants to invite 7 of them to a party. How many times will 4 particular friends never attend the party? A) 8 B) 7 C) 12 D) 15

Description :  Pramoth has 12 friends and he wants to invite 7 of them to a party. How many times will 4 particular friends never attend the party? A) 8 B) 7 C) 12 D) 15 

Last Answer : Answer: A)  Remove the 4 particular friends and invite 7 friends from the remaining 8 (12-4) friends . this can be done in 8C7 ways.  Therefore , required number of ways = = 8C7  = 8C1  = 8

If the numerator of a fraction is increased by 400% and the denominator of the fraction is increased by 300%, the resultant fraction is 18/35 . What is the original fraction? a) 72/ 175 b) 87/ 123 c) 56/232 d) 45/241

Description :  If the numerator of a fraction is increased by 400% and the denominator of the fraction is increased by 300%, the resultant fraction is 18/35 . What is the original fraction? a) 72/ 175 b) 87/ 123 c) 56/232 d) 45/241

Last Answer : Answer: A Fraction is x/y (x + 400/100 x ) / (y + 300/100 y) = 18/35 (500/100) x(35) = (400/100)y (18) (5x) × 35 = 18 × 4 y 175 x = 72 y x/y = 72/175

A wooden manufacturing company declares that a wooden is now available for $11200 as against $25200 one year before. Find the percentage reduction in the price of wooden offered by the company. a) 25 1/3 b) 691/3 c) 66 1/3 d) 33 1/3 e) 55 5/9

Description : A wooden manufacturing company declares that a wooden is now available for $11200 as against $25200 one year before. Find the percentage reduction in the price of wooden offered by the company. a) 25 1/3 b) 691/3 c) 66 1/3 d) 33 1/3 e) 55 5/9

Last Answer : Answer: E Price of the wooden a year before = $25200 Price of the wooden after a year = $11200 Decrease in price = $(25200 - 11200) = $14000 Therefore, decrease % = 14000/25200 × 100 %  = 55 5/9 %

An alloy contains 36 % of bronze. What quantity of alloy is required to get 340 g of bronze? a) 1500 b) 944.4 c) 950 d) 1000 e) 980.5

Description : An alloy contains 36 % of bronze. What quantity of alloy is required to get 340 g of bronze? a) 1500 b) 944.4 c) 950 d) 1000 e) 980.5 

Last Answer : Answer: B  Let the quantity of alloy required = x g Then 36 % of x =340 g  ⇒ 36/100 × x = 340 g  x = (340 × 100)/36 g  x = 34000/36 g x = 944.4 g

Supriya invests 35% of her monthly salary in insurance policies. She spends 45% of her monthly salary in shopping and on household expenses. She saves the remaining amount of `25,750. What is Supriya's monthly income? a) 128750 b) 160050 c) 205200 d) 263400 e) 391800

Description : Supriya invests 35% of her monthly salary in insurance policies. She spends 45% of her monthly salary in shopping and on household expenses. She saves the remaining amount of `25,750. What is Supriya's monthly income? a) 128750 b) 160050 c) 205200 d) 263400 e) 391800 

Last Answer : Answer: A  Percentage savings of Supriya = 100 – (35 + 45)  = 20%  Let her monthly income be x  x X 20 /100 = 25750  x = 25750 × 100/20  x = 128750

In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Description : In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Last Answer : Answer: A Difference = 672-126 = 546 According to the question, 70% of total aggregate = 546 Total aggregate marks = 546 × 100 /70  = 54600/70  = 780

What percent of a day in 12 hours? a) 25 b) 60 c) 45 d) 20 e) 50

Description : What percent of a day in 12 hours? a) 25 b) 60 c) 45 d) 20 e) 50

Last Answer : Answer: E  Total hours in a day = 24 Required percent = 12/24 ×100  = 50%

Find the number which when decreased by 24 % becomes 594. a) 625.8 b) 360.9 c) 465.7 d) 781.5 e) 398.5

Description : Find the number which when decreased by 24 % becomes 594. a) 625.8 b) 360.9 c) 465.7 d) 781.5 e) 398.5

Last Answer : Answer: D Let the number be m. Decrease = 24 % of m  = 24/100 × m = 6m /25 Therefore, decrease number = m – 6m/25 = (25m – 6m)/25 = 19m/25 According to the question 19m/25 = 594  19m = 594 × 25  19m = 14850  m = 14850/19  m =781.5

The population in a small town increases from 50000 to 63750 in one year. Find the percentage increase in population. a) 25 b) 62 c) 6.25 d) 27.5 e) 59.8

Description : The population in a small town increases from 50000 to 63750 in one year. Find the percentage increase in population. a) 25 b) 62 c) 6.25 d) 27.5 e) 59.8 

Last Answer : Answer: D Population in a small town last year = 50000 Population in a small town after one year = 63750 Increase in population = 63750 - 50000 = 13750 Therefore,  percentage increase in population = ... /50000 100  = 1375000/50000  = 27.5% Thus, the increase in population is 27.5%

Forty percent of Mouli’s annual salary is equal to 160% of Surya’s annual salary. Surya’s monthly salary is 80%of Gowthaman’s monthly salary. If Gowthaman’s annual salary is ` 12 lacs, what is Mouli’s monthly salary ? (At some places annual income and in some place monthly income is given.) a) 180000 b) 1200000 c) 320000 d) 250000

Description : Forty percent of Mouli's annual salary is equal to 160% of Surya's annual salary. Surya's monthly salary is 80%of Gowthaman's monthly salary. If Gowthaman's annual salary is ` 12 lacs, what is Mouli's ... income and in some place monthly income is given.) a) 180000 b) 1200000 c) 320000 d) 250000

Last Answer : Answer: C Gowthaman’s monthly salary = 12,00,000/12  = 1,00,000 Surya’s monthly salary = 1,00,000 x 80/100 = 80,000 Mouli’s monthly salary = 80,000x 1600/40  =3,20,000

A number 168 was misread as 24. Find the reading error in per cent. a) 25.7 b) 42.8 c) 42.6 d) 85.7 e) 98.6

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Last Answer : Answer: D Error = 168 – 24 = 144 Therefore, % error = 144/168 [Since, we know decrease% = decrease in value/original value × 100 %]  = 144/168 × 100  = 85.7 %

Mani went to purchase a mi mobile handset, the shopkeeper told him to pay 25% tax if he asked the bill. Mani manages to get the discount of 10% on the actual saleprice of the mobile and he paid the shopkeeper Rs.3275 without tax. Besides he manages to avoid to pay 25% tax on the already discounted price, what is the amount of discount that he has gotten?

Description : Mani went to purchase a mi mobile handset, the shopkeeper told him to pay 25% tax if he asked the bill. Mani manages to get the discount of 10% on the actual saleprice of the mobile and he paid ... to pay 25% tax on the already discounted price, what is the amount of discount that he has gotten?

Last Answer : CP = 100, SP (with tax) =125  New SP = 100 - 10 = 90  Effective discount = 125-90 = 35  So, at SP of 90 ----> discount = 35 and at SP of 3275-----> discount = 35/90*3275= Rs 1274(approx.)

In a plot of 18000 sq. m., only 13500 sq. m. is allowed for construction. What percent of the plot is to be left without construction? a) 25 b) 60 c) 45 d) 40 e) 98

Description : In a plot of 18000 sq. m., only 13500 sq. m. is allowed for construction. What percent of the plot is to be left without construction? a) 25 b) 60 c) 45 d) 40 e) 98

Last Answer : Answer: A Percentage of plot allowed for construction = (13500/18000 × 100) = 75 %. Thus, the percentage of plot to be left without construction = 100 % - 75 % = 25 %.

In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Description : In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage ... mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Last Answer : 2) 28. 57%

The difference between 45% of a number and 37% of the same number is 896. What is 25% of that number ? a) 1200 b) 2800 c) 2569 d) 3467

Description : The difference between 45% of a number and 37% of the same number is 896. What is 25% of that number ? a) 1200 b) 2800 c) 2569 d) 3467

Last Answer :  Answer: B (45 – 37)% of the number = 896  8 % of the number = 896  Number =896 x100 /8  = 89600/8  = 11200  25% of 11200 = 11200 x 25/100  = 280000 / 100  = 2800

A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Description : A man who makes a profit of 25 percent by selling sugar at Rs. 4.25/kg., lowers his price so as to gain only 5 paise per Kg. In what ratio must his sales be increased so that his total profit may be the same as before? a) 1 : 5 b) 1 : 11 c) 1 : 13 d) 1 : 17 e) 1 : 21

Last Answer : 125% C.P. of sugar per kg = Rs. 4.25 C.P of the sugar per kg = [4.25×100] / 125 = Rs. 3.40. thus, the profit on 1 kg 85 paise by reducing price to 5 paise, to make same profit he must sell, 85/5 = 17kg so required ratio = 1 : 17. Answer: d)

The number of revolutions through which a pulley will rotate from rest if its angular acceleration is increased uniformly from zero to 12 rad per sec2 during 4 sec and then uniformly decreased to 0 rad per sec2 during the next 3 sec would be a.46.4 b.26.74 c.11.6 d.3.2 e.17.7

Description : The number of revolutions through which a pulley will rotate from rest if its angular acceleration is increased uniformly from zero to 12 rad per sec2 during 4 sec and then uniformly decreased to 0 rad per sec2 during the next 3 sec would be a.46.4 b.26.74 c.11.6 d.3.2 e.17.7

Last Answer : b. 26.74

A mixture of 90 kg of rava and sugar contains 90% of sugar. The new mixture is formed by adding 30 kg of sugar. What is the percentage of sugar in the new mixture? a) 45 1/2 b) 60 1/5 c) 45 1/2 d) 92 1/2 e) 98 1/2

Description : A mixture of 90 kg of rava and sugar contains 90% of sugar. The new mixture is formed by adding 30 kg of sugar. What is the percentage of sugar in the new mixture? a) 45 1/2 b) 60 1/5 c) 45 1/2 d) 92 1/2 e) 98 1/2

Last Answer : Answer: D 90 kg = 81 kg sugar 9 kg rava Adding = 30 kg sugar Total = 111kg sugar 9kg rava (new)  111/120 × 100 = 92.5 =92 1 /2

In final exam of class IX there are 130 students 20 % students failed. How many students passed to class X? a) 105 b) 112 c) 104 d) 117 e) 104.5

Description : In final exam of class IX there are 130 students 20 % students failed. How many students passed to class X? a) 105 b) 112 c) 104 d) 117 e) 104.5

Last Answer : Answer: C) Percentage of students passed to class X = (100 % - 20 %) of 130  = 80 % of 130 =>80 % of 130 => 80/100 × 130 =>10400/100 => 104 Therefore, 104 students passed to class X

Two transformers A and B having equal outputs and voltage ratios but unequal percentage

Description : Two transformers A and B having equal outputs and voltage ratios but unequal percentage impedances of 4 and 2 are operating in parallel. The transformer A will be overloaded by 

Last Answer : Two transformers A and B having equal outputs and voltage ratios but unequal percentage impedances of 4 and 2 are operating in parallel. The transformer A will be overloaded by 33%

If the production of a factory grows at a 16% p.a., what will be its production for the year 2016 if its production in 2014 was 140 lakh tonnes? a) 222.734 b) 149.597 c) 188.384 d) 28 3.812 e) 180.534

Description : If the production of a factory grows at a 16% p.a., what will be its production for the year 2016 if its production in 2014 was 140 lakh tonnes? a) 222.734 b) 149.597 c) 188.384 d) 28 3.812 e) 180.534

Last Answer : Answer: C Required Production=140(1 + 16/100)2 lakh tones  =140(1 + 4/25)2  =140((25 + 4) /25)²  =140(29/25)²  =140(1.16)²  =140×1.3456  =188.384 lakh tonnes

The population of a town is 378000. It decreases by 16% in the 1st year and increases by 10% in the 2nd year. What is the population in the town at the end of 2 years? a) 182574 b) 482576 c) 282674 d) 283574 e) 349272

Description : The population of a town is 378000. It decreases by 16% in the 1st year and increases by 10% in the 2nd year. What is the population in the town at the end of 2 years? a) 182574 b) 482576 c) 282674 d) 283574 e) 349272

Last Answer : Answer: E After 2 years required population is  = 378000 (1-16/100) (1+10/100)  = 378000 (84/100) (110/100)  = 378000 (0.84) (1.1)  = 3,49,272

In a college election between two candidates, one candidate got 25% of the total valid votes. 30% of the votes were invalid. If the total votes were 7600, what is the number of valid votes the other candidate got ? a) 2576 b) 2314 c) 5184 d) 3990 e) 9814

Description :  In a college election between two candidates, one candidate got 25% of the total valid votes. 30% of the votes were invalid. If the total votes were 7600, what is the number of valid votes the other candidate got ? a) 2576 b) 2314 c) 5184 d) 3990 e) 9814

Last Answer : Answer: D Total valid votes = 70% of 7600 = 5320 Number of valid votes to other candidate = 75% of 5320  = 75/100 *5320 = 399000/100  = 3990

In an election three candidates received 2272 , 15272, and 23256 votes respectively. What percentage of the total votes did the winning candidate got? a) 25 b) 41 c) 75 d) 57 e) 93

Description : In an election three candidates received 2272 , 15272, and 23256 votes respectively. What percentage of the total votes did the winning candidate got? a) 25 b) 41 c) 75 d) 57 e) 93

Last Answer : Answer: D Total number of votes polled = (2272+15272+23256)  =40800 Required percentage = 23256/40800 ×100  = 2325600/40800  = 57

In an election, candidate A got 40% of the total valid votes. If 55% of the total votes were declared invalid and the total numbers of votes is 280000, find the number of valid vote polled in favour of candidate. a) 32500 b) 56000 c) 35700 d) 50400 e) 29800

Description : In an election, candidate A got 40% of the total valid votes. If 55% of the total votes were declared invalid and the total numbers of votes is 280000, find the number of valid vote polled in favour of candidate. a) 32500 b) 56000 c) 35700 d) 50400 e) 29800

Last Answer : Answer: D Total number of invalid votes = 55 % of 280000  = 55/100 280000  = 15400000/100  = 154000 Total number of valid votes 280000 - 154000 = 126000 Percentage of votes polled in favour of candidate ... favour of candidate A = 40 % of 126000  = 40/100 126000  = 5040000/100  = 50400

In an election between two candidates, one got 45% of the total valid votes, 40% of the votes were invalid. If the total number of votes was 15000, the number of valid votes that the other candidate got, was a) 2500 b) 3600 c) 4500 d) 4950 e) 9800

Description : In an election between two candidates, one got 45% of the total valid votes, 40% of the votes were invalid. If the total number of votes was 15000, the number of valid votes that the other candidate got, was a) 2500 b) 3600 c) 4500 d) 4950 e) 9800 

Last Answer : Answer: D Total number of votes = 15000 Given that 40% of Percentage votes were invalid => Valid votes = 60% Total valid votes = 15000* 60/100 1st candidate got 45% of the total valid votes. Hence ... votes that 2nd candidate got = total valid votes x 55/100  =15000*60/100*55/100  =4950

A number is reduced by 10%. Its present value is 1080. What was its original value? a) 3250 b) 2600 c) 1200 d) 2300 e) 1980

Description : A number is reduced by 10%. Its present value is 1080. What was its original value? a) 3250 b) 2600 c) 1200 d) 2300 e) 1980

Last Answer : Answer: C Original value is percentage = 100 %. Reduce amount in percentage = 10 % Therefore, Percent value in percentage = 100 % - 10 % = 90 %. According to the problem, 90 % of original value ... Therefore, 100 % of original value = 1080/100 90 = 1200. Thus, the original value was 1200.

The cost of a stencil is decreased by 30%. If the original cost is $140, find the decrease cost. a) 25 b) 68 c) 45 d) None e) 98

Description : The cost of a stencil is decreased by 30%. If the original cost is $140, find the decrease cost. a) 25 b) 68 c) 45 d) None e) 98

Last Answer : Answer: E  Original cost = $140  Decrease in it = 30% of $140  = 30/100 × 140  = 4200/100  = $42  Therefore, decrease cost = $140 - $42 = $98

By what number must the given number be multiplied to increase the number by 25 %. a) 25 b) 60 c) 50 d) None e) 98

Description : By what number must the given number be multiplied to increase the number by 25 %. a) 25 b) 60 c) 50 d) None e) 98 

Last Answer : Answer: A Let the number be m Increase in its value = 25 % of m  = 25/100 × m = m/4 Therefore, increase value = m + m/4  = (4m + m)/4  = 5m/4 Therefore, the given number must be multiplied by 5/4 to increase the number by 25 %.

In a test, minimum passing percentage for girls and boys is 60% and 25% respectively. A boy scored 560 marks and failed by 160 marks. How many more marks did a girl require to pass in the test if she scored 216 marks ? a) 2123 b) 1512 c) 2251 d) 1325 e) 3989

Description :  In a test, minimum passing percentage for girls and boys is 60% and 25% respectively. A boy scored 560 marks and failed by 160 marks. How many more marks did a girl require to pass in the test if she scored 216 marks ? a) 2123 b) 1512 c) 2251 d) 1325 e) 3989

Last Answer : Answer: B  Total marks in the test = (560 + 160) × 100 /25  = 720 ×100/25  = 72000/25  = 2880  Passing marks for girls =2880 × 60/100  = 1728  Required marks = 1728 – 216  = 1512