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Practice Test: Percentages - 1 - UPSC MCQ


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10 Questions MCQ Test CSAT Preparation - Practice Test: Percentages - 1

Practice Test: Percentages - 1 for UPSC 2024 is part of CSAT Preparation preparation. The Practice Test: Percentages - 1 questions and answers have been prepared according to the UPSC exam syllabus.The Practice Test: Percentages - 1 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Percentages - 1 below.
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Practice Test: Percentages - 1 - Question 1

Instead of a metre scale, a cloth merchant uses a faulty 120 cm scale while buying, but uses a faulty 80 cm scale while selling the same cloth. If he offers a discount of 20%, what is his overall profit percentage?

Detailed Solution for Practice Test: Percentages - 1 - Question 1

Let’s say the cost of the cloth is x rs per metre. Because of the faulty meter, he is paying x for 120 cms when buying.

So cost of 100 cms = 100x/120.

He is selling 80 cms for x, so selling price of 100cms of cloth is 100x/80.

discount = 20%

so the effective selling price is .8*100x/80= x

profit = SP-CP= x – 100x/120 = x/6

Profit % = x/6 divided by 100x/120 = 20%

Practice Test: Percentages - 1 - Question 2

Sailesh is working as a sales executive with a reputed FMCG Company in Hyderabad. As per the Company’s policy, Sailesh gets a commission of 6% on all sales upto Rs. 1,00,000 and 5% on all sales in excess of this amount. If Sailesh remits Rs. 2,65,000 to the FMCG company after deducting his commission, his total sales were worth:

Detailed Solution for Practice Test: Percentages - 1 - Question 2

Let total sales be ‘x’

The commission that Sailesh will get is x – 265000

He gets 6% on sales upto 100000 and 5% on sales greater than that.

Calculating his commission on total sales:

0.06*100000 + 0.05(x-100000)

Equating,

0.05x + 1000 = x – 265000

0.95x = 266000

x= 280000

Hence, his sales were worth 280,000

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Practice Test: Percentages - 1 - Question 3

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

Detailed Solution for Practice Test: Percentages - 1 - Question 3

Let the number of girls be 2x and number of boys be x.

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x – 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

Practice Test: Percentages - 1 - Question 4

The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

Detailed Solution for Practice Test: Percentages - 1 - Question 4

The ratio of number of male and female journalists in a newspaper office is 5:4.

The newspaper has two sections, political and sports.

If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

Let ‘9x’ be the number of total journalists in the office.
Then, we can say that the number of male and female journalists are ‘5x’ and ‘4x’ respectively.

It is given that 30 percent of the male journalists and 40 percent of the female journalists are covering political news. Hence, total number of journalists who are covering political news = 0.3*5x + 0.4*4x = 3.1x

Therefore, the total number journalists who are covering sports news = 9x – 3.1x = 5.9x.
Hence, the percentage of the journalists in the newspaper is currently involved in sports reporting = 5.9x/9x x 100 ≈ 

65 percent. Therefore, option B is the correct answer.

Practice Test: Percentages - 1 - Question 5

If equal numbers of people are born on each day, find the approximate percentage of the people whose birthday will fall on 29th February. If we are to consider people born in 20th century (1901-2000) and assuming no deaths.

Detailed Solution for Practice Test: Percentages - 1 - Question 5

Assume one person is born every day. In 100 years, there will be 25 leap years. So 25*1 additional people will be born on these days.
So, total people born will be = 365 x 100 x 1 + 25 x 1
And people born on 29th february = 25 x 1
Hence percentage will be 

Practice Test: Percentages - 1 - Question 6

In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to

Detailed Solution for Practice Test: Percentages - 1 - Question 6

Let the total number of people in the group be x

So, the total number of young members = (28/100) × x = 0.28x

Also, the total number of old members = x – 0.28x = 0.72x

It is also given that, 65% of all the members are literates,

Hence, total number of literate people = (65/100) × x = 0.65x

So, the total number of illiterate people = x – 0.65x = 0.35x

Besides, it is also given that 25% of the literates are young.

So, the number of young literates = (25/100) × 0.65x = 0.1625x

Hence, the number of young illiterates = 0.28x – 0.1625x = 0.1175x

Also, we get:

The number of illiterate old people = 0.35x – 0.1175x = 0.2325x

Hence, we can obtain the percentage of old people among the illiterates as:

(0.2325x/0.35x) × 100 = 66.423% ≈ 66%

∴ The percentage of old people among the illiterates is nearly 66% 

Practice Test: Percentages - 1 - Question 7

Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

Detailed Solution for Practice Test: Percentages - 1 - Question 7

Let the CP of the each toy be “x”. CP of 12 toys will be “12x”. Now the shopkeeper made a 10% profit on CP. This means that

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be “m”. Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m  Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

Practice Test: Percentages - 1 - Question 8

A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance, he purchases oranges. Number of oranges he can purchase is:

Detailed Solution for Practice Test: Percentages - 1 - Question 8

The person can buy 50 oranges or 40 mangoes.

Let the price of one orange be Rs. x

Total amount the person has = Rs. 50x

40 mangoes cost 50x, So one mango costs 1.25x

10% of the total amount is retained for taxi fare = 10% of 50x = 5x

20 mangoes bought for 20 x 1.25x = 25x

Money left with the person = 50x – (Taxi fare) – (Mangoes cost)

= 50x – 5x – 25= 20x

One Orange was for Rs. x, Therefore, 20 oranges can be bought with Rs. 20 x

Thus, the person bought 20 oranges.

Practice Test: Percentages - 1 - Question 9

A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is reduced onto sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closest to:

Detailed Solution for Practice Test: Percentages - 1 - Question 9

No. of Characters in one line = 65

No. of characters in one sheet = No. of lines × No. of characters per line = 55 × 65

Total number of characters = No. of sheets × No. of characters in one sheet = 20 × 55 × 65 = 71500

If the report is retyped –

New sheets have 65 lines, with 70 characters per line

No. of characters in one sheet = 65 × 70

Number of pages required,

Hence, 16 pages will be required if report is retyped.

Hence, reduction of (20 – 16) = 4 pages

% reduction is = (4/20) x 100 = 20%

Practice Test: Percentages - 1 - Question 10

Seema has joined a new Company after the completion of her B.Tech from a reputed engineering college in Chennai. She saves 10% of her income in each of the first three months of her service and for every subsequent month, her savings are Rs. 50 more than the savings of the immediate previous month. If her joining income was Rs. 3000, her total savings from the start of the service will be Rs. 11400 in:

Detailed Solution for Practice Test: Percentages - 1 - Question 10

Seema saved Rs. 900 in the first 3 months. She must saved Rs. (11400 – 900) = Rs. 10500 in the subsequent months.
The sequence will be of the form: 350 + 400 +……….. n terms = 10500

Solving, we get n = 15
The savings of Rs. 10500 is done in 15 months. Seema saved Rs. 11400 in 15+3 = 18 months.
Hence, option A is the correct answer.

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