24-08-2021 Answers

 1. A student multiplies a number by 9/5 instead of multiplying it by 5/9. What is the percentage change in the resulting value due to this mistake?

Answer: None of these

Solution: Let the number be n.

Expected result = 5n/9

Obtained result = 9n/5

Difference = 9n/5 - 5n/9 = 56n/45

Percentage change = 100 * (56n/45)/(5n/9) = 224%

2. A dosa shop increased the price of a dosa from Rs.20 to Rs.25. By what percentage should a customer decrease his dosa consumption to keep the food expense constant?

Answer: 20%

Solution: Increase in price ratio = x = (25-20)/20 = 1/4.

Hence decrease in consumption ratio = y = x/(1+x) = 1/4/(1+1/4)= 1/5.

Let original food expense be F.

% decrease = 100*(F/5)/F = 20%.

3. 5000 voted in an election between two candidates. 14% of the votes were invalid. The winner won by a margin approximately closer to 15%. Find the number of votes secured by the winner.

Answer: 2473

Solution: Valid votes = 5000 * 86/100 = 50*86 = 4300

Let num of votes got by winner = w.

Votes got by the other person = 4300 - w.

w - (4300 -w) = 0.15 * 4300,

w=2472.5 which is approximately 2473.

4. Fresh tomato contain 60% water. If 10 kgs of dry tomato can be obtained from 20 kgs of fresh tomato, what is the percentage of water in dry tomato?

Answer: 20%

Solution: Tomato matter in fresh tomato = 40%.

This 40% remains constant in fresh and dry tomato.

Hence 40% weight of fresh tomato of 20kgs = 8kgs of tomato matter.

So the 10kgs of dry water melon has 8kgs of tomato matter and 2kgs of water.

So % of water = 2/10 * 100 = 20%. 

5. A single discount equivalent to three successive discounts of 20%, 25% and 10% is

Answer: 46%

Solution: Let the initial price be x.

After first discount it is x*0.8 = 0.8x.

After second discount it is 0.8x * 0.75.

After third discount it is 0.8x * 0.75 * 0.9 = 0.72x * 0.75 = 0.54x = 54% of x.

Hence discount = 100 - 54 = 46%

6. Janakiraman lost his job, which reduced the net family income by 20%. His wife decided to work overtime to compensate. By what percentage must her net salary increase in order to arrive at the same net family income as they had before Janakiraman lost his job?

Answer: 25%

Solution: Let original net income be x.

Janakiraman's salary was 0.2x (as reduction by 20%).

Wife's salary was 0.8x.

Now additional pay for wife should be 0.2x.

Hence percentage hike required in wife's salary =(0.2x/0.8x)*100 = 25%.

7. If 20% of A = B and 40% of B = C, then 60% of (A + B) is

Answer: None of these

Solution: From the given data, A=5B and 2B/5=C.

Hence A+B = 6B.

so 60% of (A+B) = 6/10 * 6B = 36B/10.

As C=2B/5, expressing 60% of (A+B) as a % of C, it is 100 * (36B/10)/(2B/5) = = 900%.

Hence answer is "None of the above"

8. After three successive equal percentage rise in the salary, the sum of 100 rupees turned into 140 rupees and 49 paise. Find the percentage rise in the salary each time.

Answer: 12%

Solution: Let the % hike be p.

(1+p)*(1+p)*(1+p)*100 = 140.49

(1+p)^3 = 1.4049,

Now solving using given options, nearest value is 12%

9. After 76 percent of birds flew from one sanctuary to another, the number of birds is reduced to 78. What is three fifths of the number of birds that were initially there in the sanctuary?

Answer: 195

Solution: Let the original number be x.

x-0.76x=78, x=78/0.24

 

Three-fifth = 3x/5 = 3/5 * 78/0.24 = 195

10. In a college 30% of the stduents belong to ECE, 40% of the rest belong to EEE and 50% of the remaining belong to Mechanical departments. The remaining 210 students belong to MBA. What is the total number of students in the college.

Answer: 1000

Solution: Answer is 1000 students.

Let the total num of students in the college be x.

ECE students = 0.3x

EEE students = (x-0.3x)*0.4 = 0.7x*0.4 = 0.28x

Mech students = (x-0.3x-0.28x)*0.5 = 0.42x * 0.5 = 0.21x

x - 0.3x - 0.28x - 0.21x = 210.

x = 210/0.21 = 1000.

11. Satish bought a microwave oven and paid 20% less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did Satish earn on the original price?

Answer: 4%

Solution: Let original price be x.

C.P=0.8x

Selling price = 0.8x * 1.3 = 1.04x.

Profit % = 100 * (1.04x - x)/x % =4%

12. Marked price of an item is 60% of it's MRP. A person buys the item at half it's marked price. What is the percentage of discount given?

Answer: 70%

Solution: Let MRP be x.

Marked price =0.6x.

Purchased price = Half of 0.6x = 0.3x.

so discount = x-0.3x = 0.7x which is 70%.

13. A ball gets dropped from a height of 100842 cms. Every time it bounces it reaches a height which is 14.28% less than the previous height reached. What height will the ball reach on its fifth bounce?

Answer: 46656 cms

Solution: 0.1428 is nothing but 1/7.

 

Hence each time the ball bounces 1/7th of the previous height reached is lost. That is the ball reached 6/7th of the previous height.

 

Hence height reached on the fifth bounce = 100842 * (6/7) * (6/7) * (6/7) * (6/7) * (6/7) = 6*6*6*6*6*6 = 46656 cms.

14. As the price of gold increased by 40%, 2 grams of less gold is available for Rs.28000. What was the original price of gold per gram?

Answer: None of these

Solution: Ans: None of the above (As answer is Rs.4000)

Let initially x grams of gold be brought for Rs.28000.

Original price = 28000/x per gram.

 

 

Now due to 40% hike, (x-2) grams available for Rs.24000

Hence 1.4 * 28000/x * (x-2) = 28000,

Solving x=7

 

 

Hence original price per gram = 28000/7 = Rs.4000

15. In 2008, one kilogram of fruit was sold at Rs.25. If the fruit rate increases at 2.5% more than the inflation rate which is 5.5% a year, what will be the cost of one kilogram of fruit after 2 years?

Answer: Rs 29.16

Solution: Rate of increase per year = 5.5%+2.5% = 8%.

Hence after 1 year price of fruit = 25 * 1.08 = Rs.27

After 2 years price = 27 * 1.08 = Rs. 29.16

16. Due to a 10% hike in petrol price Balu's fuel expense increased by 5%. By what percentage did he reduce the consumption of petrol?

Answer: 4.54%

Solution: Even though the petrol hike was 10%, Balu reduced the consumption and brought the expense down.

Hence reduction in consumption = 100 * 5/110 = 4.54%

17. 500 students appeared in an exam. 70% of the girls and 75% of the boys passed the exam. If the total pass % is 72, find the number of girls who appeared for the exam.

Answer: 300

Solution: Mean = m = 72, low = a = 70, high = x = 75.

Num of girls/Num of boys = (x-m)/(m-a) = 3/2.

So let the num of girls = 3x and boys be 2x.

3x+2x = 5x = 500, x=100.

So num of girls who appeared for exam = 3x = 300.

18. Of the applicants who passed a certain test, 30 applied for both college X and Y. If 20% of the applicants who applied for college X and 25% of the applicants who applied for college Y, applied for both colleges X and Y, how many applicants applied for only one of the colleges?

Answer: 210

Solution: Let x be who applied only for college X and y who applied only for college Y.

Then 0.2(x+30) = 30 and 0.25(y+30) = 30.

So x = 120 and y = 90 so x+y = 210.

19. The salaries of Anita and Banu are 40% and 50% more than the salary of Chitra. Approximately by what percentage is Anita's salary less than Banu's?

Answer: 6.66%

Solution: Let Chitra's salary be x.

Anita's salary = 1.4x and Banu's = 1.5x

Difference = 1.5x - 1.4x = 0.1x

% lesser = 100 * 0.1x/1.5x = 100/15 % = 6.66% approx

20. There are 3 shop keepers A,B,C. A sells goods 25% cheaper than B but 25% more dearer than C. By how much cheaper does C sell when compared to B?

Answer: 40% cheaper

Solution: Let B sell at x price. 

A sells at 0.75x.

Let C sell at y price. 

 

From given data, 0.75x = 1.25y,

y = 0.75x/1.25 = 3x/5.

Hence C sells 2x/5 lesser which is 40% lesser than B.