PERCENTAGES EXAMPLE QUESTIONS PART 2
1. Raman's salary was decreased by 50% and subsequently increased by 50%. How much percent does he loss.
A. 75
B. 65
C. 45
D. 25
Answer And Explanation
Answer: Option D
Explanation:
Let the origianl salary = Rs. 100
It will be 150% of (50% of 100)
= (150/100) * (50/100) * 100 = 75
So New salary is 75, It means his loss is 25%
2. If x% of y is 100 and y% of z is 200, then find the relation between x and z.
A. z = x
B. 2z = x
C. z = 2x
D. None of above
Answer And Explanation
Answer: Option C
Explanation:
It is , y% of z = 2(x% of y)
=> yz/100 = 2xy/100
=> z = 2x
3. Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first
A. 8%
B. 9%
C. 10%
D. 11%
Answer And Explanation
Answer: Option C
Explanation:
Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10
Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what percent of first number
=> (7x/100 * 10/7x * 100 )% = 10%
4. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate got
A. 55%
B. 56%
C. 57%
D. 58%
Answer And Explanation
Answer: Option C
Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
5. A reduction of 21% in the price of an item enables a person to buy 3 kg more for 100. Thereduced price of item per kg is:
(a) Rs. 5.50
(b) Rs. 7.50
(c) Rs. 10.50
(d) Rs. 7.00
Solution:(d)
Reduced price will be:
Rp/100y per kg
In our case R= Rs. 100 , x=21% , y=3kg
{(100 x 21)/(100 x 3)} = Rs. 7
6. A vessel has 60 L of solution of acid and water having 80% acid. How much water is to be added to make it solution in which acid forms 60%?
(a) 48 L
(b) 20 L
(c) 36 L
(d) None of these
Solution: (b)
Given, percentage of acid = 80%
Then, percentage of water = 20%
In 60L of solution, water = (60 x 20)/ 100 = 12L
Let p liter of water is to be added.
According to the question,
=>{(12 + p)/(60 + p)} x 100 = 40 (∵ 100 – 60 = 40% water)
=>1200 + 100p = 2400 + 40p
⇒ 60p = 1200
p= 20L
7. If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction become 5/2. The original fraction is:
(a) 24/19
(b) 3/18
(c) 95/48
(d) 48/95
Solution: (c)
Let original fraction be p/y
According to the question,
{(120/100)p/(95/100)y} = 5/2
120p/95y = 5/2
=> p/y = (5/2)x (95/120) = 95/48.
8. The monthly income of a person was Rs 13500 and his monthly expenditure was Rs 9000. Nextyear his income increased by 14% and his expenditure increased by 7%. The per cent increase inhis savings was:
(a) 7%
(b) 21%
(c) 28%
(d) 35%
Solution: (c)
Given, monthly income = 13500 and expenditure = 9000
Then, original savings= Rs. (13500-9000) = Rs 4500
New income = 114% of Rs. 13500 = Rs 15390
New expenditure= 107% of Rs 9000 = Rs 9630
New saving = Rs. (15390 – 9630) = Rs 5760
NS = new savings
OS = Original savings
Percentage increase in savings = {(NS – OS)/OS} X 100
{(5760 – 4500)/4500} X 100 = (1260/4500)X 100 = 28%
9. In a quarterly examination a student secured 30% marks and failed by 12 marks. In the same examination another student secured 40% marks and got 28 marks more than minimum marks to pass. The maximum marks in the examination is:
(a) 300
(b) 500
(c) 700
(d) 400
Solution : (d)
here a=12, b=28, x=30%, y= 40%
max marks = 100(a+b)/(x-y)
4000/10 =400
10. In an examination, 52% of the candidates failed in English and 42% failed in Mathematics. If17% failed in both the subjects, then the percentage of candidates who passed in both the subjects, was:
(a) 23%
(b) 21%
(c) 25%
(d) 22%
Solution: (a)
In an examination a% of total number of candidates failed in a subject X and b% of total number of
candidates failed in subject Y and c% failed in both subjects.
Then, percentage of candidates who passed in both the subjects: [100–(a + b– c)%].
Why so?
We need to subtract failed candidate from total to calculate pass candidates, but in this process, we subtract people who were fail in both of subjects twice, so we added c again.
Now equation can be read as [ 100—a-b+c]
Here, a = 52, b = 42 and c = 17, then Percentage of candidates who passed in both the subjects
= [100 –(52 +42–17)]
= 100–77=23%
11. lf A has 4/5 of the number of books that shelf B has. If 25% of the books A are transferred to B and then 25 % of the books from B are transferred to A then the percentage of the total number of books that on shelf A is:
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Solution: (b)
Let the number of books in shelf B be 100.
∴ Number of books in shelf A = {(100 x 4)/5} = 80
On transferring 25% i.e.,of books of shelf A to shelf B, the books on
shelf B = {100 + (80 x 25)/100}
B = 100 + 20 = 120
Books left in shelf A = 80-20 = 60
Again, on transferring 1/4th books of shelf B to shelf A, the books on
shelf A = {60 + (120/4)} = 90
Total no of books in A and B = 120 +60 = 180
Required percentage of books on shelf A = (90/180) x 100 = 50%
12. Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calcualte average marks of the remaining students of the class.
A. 60
B. 70
C. 75
D. 80
Answer And Explanation
Answer: Option C
Explanation:
Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
=> 70x = 8000 - (950 + 1800) = 5250
=> x = 75.
13. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are
A. 42,30
B. 42,31
C. 42,32
D. 42,33
Answer And Explanation
Answer: Option D
Explanation:
Let their marks be (x+9) and x.
Then, x+9 = 56/100(x + 9 +x)
=> 25(x+9)
=> 14 (2x + 9)
=> 3x = 99
=> x = 33.
So, their marks are 42 and 33
14. One fourth of one third of two fifth of a number is 15. What will be40% of that number
A. 140
B. 150
C. 180
D. 200
Answer And Explanation
Answer: Option C
Explanation:
(1/4) * (1/3) * (2/5) * x = 15 then x = 15 * 30 = 450
40% of 450 = 180
15. 10% of inhabitants of a village having died of cholera, a panic set in, during which 25% of the remaining inhabitants let the village. The population is then reduced to 4050. Find the original inhabitants
A. 5500
B. 6000
C. 6500
D. 7000
Answer And Explanation
Answer: Option B
Explanation:
Let the total number is x,
then,
(100-25)% of (100 - 10)% x = 4050
=> 75% of 90% of x = 4050
=> 75/100 * 90/100 * x = 4050
=> x = (4050*50)/27 = 6000
16. What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A.
1
B.
14
C.
20
D.
21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = 14 x 100 % = 20%.
70
17.
What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
A.
1
B.
14
C.
20
D.
21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = 14 x 100 % = 20%.
70
18.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
A.
2700
B.
2900
C.
3000
D.
3100
Answer: Option A
Explanation:
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (0.45 x 6000) = 2700.
19.
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A.
Rs. 200
B.
Rs. 250
C.
Rs. 300
D.
None of these
Answer: Option B
Explanation:
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
z + 1.20z = 550
11/5z = 550
z = 550/11 x 5 = 250.
20.
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
A.
4.37%
B.
5%
C.
6%
D.
8.75%
Answer: Option B
Explanation:
Increase in 10 years = (262500 - 175000) = 87500.
Increase% = 87500/175000 x 100 % = 50%.
Required average = 50/10 % = 5%.
21.
Two persons Raj and Ramu started working for a company in similar jobs on January 1, 1991. Raj's initial monthly salary was Rs 400, which increases by Rs 40 after every year. Ramu's initial monthly salary was Rs 500 which increases by Rs 20 after every six months. If these arrangements continue till December 31, 200. Find the total salary they received during that period.
A. Rs 1,08,000
B. Rs 1,44,000
C. Rs 1,32,000
D. Rs 1,52,400
Solution:
Option(D) is correct
Raj's salary as on 1 jan 1991 is Rs 400 per month
His increment in his month salary is Rs 40 per annum
His total salary from 1 jan 1991 to 31st dec 2000
i.e. in ten years
=12[2(400)+(10−1)40]×102
=Rs 69,600
Ramu's salary as on Jan 1st 1991 is Rs 550 and his half yearly increment in his month salary is Rs 20.
His total salary from 1 jan 1991 to dec 31, 2000
=6[2(500)+(20−1)20]×202
=Rs 82,000
Total salary of Raj and Ramu in the ten year period:
= Rs. 69600+ Rs. 82800
⇒ Rs 1,52,400
22.
30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 15%
B. 20%
C. 80%
D. 70%
Solution:
Option(C) is correct
20% of the men are above the age of 50 years. 20% of these men play football. Therefore, 20% of 20% or 4% of the total men are football players above the age of 50 years.
20% of the men are football players. Therefore, 16% of the men are football players below the age of 50 years.
Therefore, the % of men who are football players and below the age of
50=16/20×100
23.
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6, then what was the cost of the tax free items?
A. Rs. 15
B. Rs. 15.70
C. Rs. 19.70
D. Rs. 20
Solution:
Option(C) is correct
Let the amount taxable purchases be Rs. x
Then,
6% of x=30100
x=30100×1006
=5
Cost of tax free items
= Rs. [25−(5+0.30)]
= Rs. 19.70
24.
If the cost price of 20 articles is equal to the selling price of 16 articles, What is the percentage of profit or loss that the merchant makes?
A. 20%Profit
B. 25% Loss
C. 25% Profit
D. 33.33% Loss
Solution:
Option(C) is correct
Let Cost price of 1 article be Re.1.
Therefore, Cost price of 20 articles = Rs. 20.
Selling price of 16 articles = Rs. 20
Therefore, Selling price of 20 articles
=2016×20
=25
Profit = Selling price - Cost price
=25−20=5
Percentage of profit=profitC.P×100
=520×100=520×100
= 25% Profit
25.
A man purchased a bag for Rs.360 and sold it the same day for Rs. 360, allowing the buyer a credit of 9 years.
If the interest be 712% then what is the man's gain?
A. 159
B. 173
C. 243
D. 287
Solution:
Option(C) is correct
Interest rate =152=7.5%
CP=Rs. 360.
SP=360+intrest on 360 for 9 years
Interest = 360×15×92×100
=Rs. 243.
Gain = Interest =Rs. 243.
26.
Two merchants sell, each an article for Rs.1000. If Merchant AA computes his profit on cost price, while Merchant BB computes his profit on selling price, they end up making profits of 25% respectively.
By how much is the profit made by Merchant BB greater than that of Merchant AA?
A. Rs.50.00
B. Rs.66.67
C. Rs.75.00
D. Rs.150.00
Solution:
Option(A) is correct
Merchant B computes his profit as a percentage of selling price. He makes a profit of 25% on selling price of Rs.1000. i.e. his profit =25/1000
Merchant A computes his profit as a percentage of cost price.
If he makes a profit of 25% or 1/4th of his cost price, then his profit expressed as a percentage of selling price,
=11+4=(15)th or 20% of selling price.
So, Merchant A makes a profit of 20% of Rs. 1000=Rs. 200
Hence, Merchant B makes Rs.50Rs.50 more profit than Merchant A.
27.
After offering a discount of 37.5, Pankaj sold the rice at a profit of 25. Had he offered a discount of 41.67, his profit or loss percent would have been:
A. 16.66% profit
B. 12% loss
C. 29.17% loss
D. 8.33% profit
Solution:
Option(A) is correct
Let the marked price of the rice = 8p
Discount
=37.5%=38×MP
Selling price
=8p−38×8p=5p
5p=CP+25100×25
⇒CP=4p
If he had offered a discount of 41.67%=512
SP=8p−512×8p
=14p3
Profit
=14p3−4p
=2p3
Profit percentage
=2p/34p×100
=1006%
=16.66%
28.
The cost of raw material of a product increases by 30, the manufacturing cost increases by 40 and the selling price of the product increases by 60. The raw material and the manufacturing cost, originally, formed 40 and 60 of the total cost respectively. If the original profit was one-fourth the original manufacturing cost, find the approximation new profit percentage.
A. 48.39%
B. 54.68%
C. 62.48%
D. 42.36%
Solution:
Option(A) is correct
Let the total initial cost of the product be Rs 100.
Manufacturing cost = Rs 60
Raw materials cost = Rs 40
Also, original selling price
= Rs100+25% of 60=Rs 115
New raw material cost
= Rs 40+30% of 40= Rs 52
New manufacturing cost
== Rs 60+20% of 60= Rs 72
New cost of the product = Rs124
New selling price
=115+60% of 115=Rs 184
New profit percentage
=60124×100
=48.38%
29.
In a period from January to March, Jamshedpur Electronics sold 3150 units of Television, having started with a beginning inventory of 2520 units and ending with an inventory of 2880. What was the value of order placed (Rupees in thousands) by Jamshedpur Electronics during the three months period? [Profits are 2525 of cost price, uniformly.]
A. 26325
B. 22320
C. 25200
D. 28080
Solution:
Option(D) is correct
Units ordered = Units sold + Ending Inventory-Beginning Inventory
⇒3150+2880−2520=3510
Total sales of Television in Rs. Thousand
⇒900+1800+6300+1050+2100+7350+1200+2400+8400=31500
Sales Price per unit of Television in Rs. Thousand
= 31500/3150=10
Profits are 25%of the cost price.
Sales Price = Cost Price + Profits = Cost Price +0.25×+0.25× Cost Price =1.25×=1.25× Cost Price
Cost Price per unit of Television
=Sales Price per unit1.25
=101.25
=8
The value of the order placed in Rs. Thousand = Units ordered ×× Cost Price per unit
⇒3510×8=28080
30.
A Techno company has 14 machines of equal efficiency in its factory. The annual manufacturing costs are Rs 42,000 and establishment charges are Rs 12,000. The annual output of the company is Rs 70,000.
The annual output and manufacturing costs are directly proportional to the number of machines. The shareholders get 12.5 profit, which is directly proportional to the annual output of the company.
If 7.14 machines remain closed throughout the year, then the percentage decrease in the amount of profit of the shareholders would be:
A. 12%
B. 12.5%
C. 13.5%
D. None of these
Solution:
Option(B) is correct
Original profit
= 70,000−42,000−12,000=16,000
If 7.14% of 14 i.e. one of the machines closed throughout the year, then change in profit will be:
=1314×[70,000−42,000]
=14,000
Thus, decrease in the profit %%
=2000/16000×100
=12.5%
31.
If the price of petrol increases by 25 and Raj intends to spend only an additional 15 on petrol, by how much will he reduce the quantity of petrol purchased?
A. 10%
B. 12%
C. 8%
D. 6.67%
Solution:
Option(C) is correct
Let the price of 1 litre of petrol be Rs. x and let Raj initially buys 'y litres of petrol.
Therefore, he would have spent Rs. xy on petrol.
When the price of petrol increases by 25%, the new price per litre of petrol is 1.25x.
Raj intends to increase the amount he spends on petrol by 15%
i.e., he is willing to spend xy+15% of xy=1.15xy
Let the new quantity of petrol that he can get be 'q.
Then, 1.25x×q=1.15xy
q=1.15xy1.25x
=1.15y1.25
=0.92y
As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what he used to get earlier.
Or a reduction of 8%
32.
A shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by xx during the year. A famine hits his village in the next year and many of his sheep die. The sheep population decreases by yy during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheep. Which of the following is correct?
A. x>y
B. y>x
C. x=y
D. Cannot be determined
Solution:
Option(A) is correct
Let us assume the value of x to be 10%
Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million
In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.
i.e., 0.1 million sheep have died in 2001.
In terms of the percentage of the number of sheep alive at the beginning of 2001,
it will be (0.1/1.1)×100%=9.09%(0.1/1.1)×100%=9.09%.
From the above illustration it is clear that x>y
33.
Peter got 30 of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40 of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
A. 35
B. 250
C. 75
D. 85
Solution:
Option(D) is correct
Let x be the maximum marks in the examination.
Therefore, Peter got 30% of x
=30100×x=0.3x
And Paul got 40% of x
=40100×x
=0.4x
In terms of the maximum marks Paul got 0.4x−0.3x=0.1x more than Peter. --------(1)
The problem however, states that Paul got 15 marks more than the passing mark and Peter got 10 marks less than the passing mark.
Therefore, Paul has got 15 + 10 = 25 marks more than Peter.-------- (2)
Equating (1) and (2), we get
0.1x=25
⇒x=250
xx is the maximum mark and is equal to 250 marks.
We know that Peter got 30% of the maximum marks.
Therefore, Peter got
=30100×250
=75 marks
We also know that Peter got 10 marks less than the passing mark.
Therefore, the passing mark will be 10 marks more than what Peter got = 75+10=85
34.
In an acoustics class, 120 students are male and 100 students are female. 25% of the male students and 20% of the female students are engineering students.
20% of the male engineering students and 25% of the female engineering students passed the final exam.
What percentage of engineering students passed the exam?
A. 5%
B. 10%
C. 16%
D. 22%
Solution:
Option(D) is correct
There are 100 female students in the class, and 20% of them are Engineering students.
Now, 20% of 100 equals (20100)×100=20..
Hence, the number of female engineering students in the class is 20.
Now, 25% of the female engineering students passed the final exam:
25.
Hence, the number of female engineering students who passed is 5.
There are 120 male students in the class. And 25% of them are engineering students.
Now, 25% of 120 equals(25100)×120=(14)×120=30
Hence, the number of male engineering students is 30.
Now, 20% of the male engineering students passed the final exam:
20
Hence, the number of male engineering students who passed is 6.
Hence, the total number of engineering students who passed is:the total number of engineering students who passed is:
(Female Engineering students who passed)+(Female Engineering students who passed)+
(Male Engineering students who passed)(Male Engineering students who passed)
=5+6=11
The total number of engineering students in the class is:The total number of engineering students in the class is:
(Number of female engineering students)+(Number of female engineering students)+
(Number of male engineering students)=(Number of male engineering students)=
=30+20=50
Hence, the percentage of engineering students who passed is
(Total number of engineering students who passedTotal number of engineering students)×100(Total number of engineering students who passedTotal number of engineering students)×100
=(1150)×100
= 22%
35.
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 45%
B. 45 5/11%
C. 54 6/11%
D. 55%
Solution:
Option(B) is correct
Number of runs made by running,
=110−(3×4+8×6)=110−(3×4+8×6)
=110−(60)=110−(60)
=50.=50.
Required percentage,
=(50110×100)%=(50110×100)%
=45 5/11%
36.
If bb equals 10% of a and c equals 20% of b, then which one of the following equals 30% of c?
A. 0.006% of a
B. 0.006% of a
C. 0.06% of a
D. 0.6% of a
Solution:
Option(D) is correct
b=10% of a=(10100)×a=0.1aa=(10100)×a=0.1a
c =20% of b= =(20/100)×b
=0.2b=0.2×0.1a=0.2b=0.2×0.1a
Now, 30% of c=(30/100)×c
=0.3c=(0.3)(0.2)(0.1a)
=0.6%a
37.
8 is 4% of a, and 4 is 8% of b. c equals b/a. What is the value of c?
A. 1/32
B. ¼
C. 1
D. 4
Solution:
Option(B) is correct
4% of a is 4a/100.
Since this equals 8, we have 4a/100=8.
Solving for a yields a=8×(100/4)=200.
Also, 8% of bb equals 8b/100 and this equals 4.
Hence, we have (8/100)×b=4.
Solving for b yields b=50
Now, c=ba
=50/200
=1/4.
38.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid.
If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
A. 2700
B. 2900
C. 3000
D. 3100
Solution:
Option(A) is correct
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
=(45/100)×6000 = 2700.
39.
If 50% of xx equals the sum of yy and 20, then what is the value of x–2yx–2y?
A. 20
B. 40
C. 60
D. 80
Solution:
Option(B) is correct
50% of xx equals the sum of yy and 20. Expressing this as an equation yields:
(50/100)×x=y+20
x/2=y+20
x=2y+40
x–2y= 40
40.
The price of an item changed from Rs. 120 to Rs. 100. Then later the price decreased again from Rs. 100 to Rs. 80.
Which can be said about the two decreases in percentage term?
A. 1st decrease is larger
B. 2nd decrease is larger
C. both are equal
D. nothing can be said
Solution:
Option(B) is correct
First decrease in percent,
partwhole=(120−100)120
=0.17=17%
Second decrease in percent,
partwhole=(100−80)100
=0.20=20%
The second decrease is larger in percent term.
The parts were the same in both cases, but the whole was smaller in the second decrease.
41.
Sahil earns 10 percent more than Satish and satish earns 20 percent more than Swati. if swati earns rs 17,500 less than Sahil, what is the earnings of each?
A. Swati = Rs. 35,900.0, Sahil = Rs. 53,400.0, Satish = Rs. 58,740.0
B. Swati = Rs. 39,500.0, Sahil = Rs. 57,000.0, Satish = Rs. 62,700.0
C. Swati = Rs. 54,687.5, Sahil = Rs. 72,187.5, Satish = Rs. 65,625.0
D. Swati = Rs. 69125.0, Sahil = Rs. 86,625.0, Satish = Rs. 95,287.5
Solution:
Option(C) is correct
Let Swati earns Rs. x.
Satish earns =x+20/100
Sahil earns =1.2x+10/100
Earning difference of Sahil and Swati =17,500
1.32x−x=17,500
0.32x=17,500
⇒x=175000.32
⇒x=Rs. 54,687.5
So, Swati's earnings,
x=Rs. 54,687.5
Sahil's earnings,
=1.32×x
=1.32×Rs. 54,687.5
=Rs. 72,187.5
Satish's earnings,
=1.2×x
=1.2×Rs. 54,687.5
=Rs. 65,625.0
42.
A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest.
What percent of his apples does the vendor throw?
A. 17
B. 23
C. 77
D. None of these
Solution:
Option(B) is correct
Let the number of apples be 100.
On the first day he sells 60% apples i.e.,60 apples. Remaining apples =40.
He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has 40−6=3440−6=34 apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in total he throws 6+17=6+17= 23 apples.
43.Fresh grapes contain 60% water, while dry grapes contain 40% water. How many kgs of dry grapes can be obtained by 60 kg of fresh grapes?
A) 25 kgs B) 30 kgs
C) 35 kgs D) 40 kgs
Answer & ExplanationAnswer: D) 40 kgs
Explanation:
The fruit content in both the fresh fruit and dry fruit is the same.
Let 'p' kgs of dry grapes are obtained from the 60 kgs of fresh grapes
=> 60 x 40/100 = p x 60/100=> p = 40 kgs.
Therefore, 40 kgs of Dry grapes are obtained from 60 kgs of Fresh grapes.
44.In Veeru Bhai Pvt.limited company 60% of the employees are men and 48% of the employees are Engineer and 66.6% of these are men. The percentage of women who are not engineers:
A) 33.33% B) 60%
C) 52% D) 46.66%
Answer & ExplanationAnswer: B) 60%
Explanation:
Men Women
600x 400x
total engineers = 480x
Male engineers = 480x X 0.66 = 320
Women who are Engineers = 160x
Women are not Engineers = 400x - 160x = 240x
Required percentage = 240/400x100 = 60%
45.A man lost half of its initial amount in the gambling after playing 3 rounds. The rule of gambling is that if he wins he will receive Rs. 100, but he has to give 50% of the total amount after each round.Luckily he won all the three rounds. The initial amount with which he had started the gambling was :
A) 500/3 B) 700/3
C) 300 D) 600
Answer & ExplanationAnswer: B) 700/3
Explanation:
Let the initial amount be x (with gambler), then
[[(x+100)1/2 + 100]12+100]1/2 = x/2
=> x = 700/3
46.The Shopkeeper increased the price of a product by 25% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 70% of the required amount. What is the net difference in the expenditure on that product?
A) 10% more B) 5% more
C) 12.5% more D) 17.5% more
Answer & ExplanationAnswer: C) 12.5% more
Explanation:
Quantity X Rate = Price
1 x 1 = 1
0.7 x 1.25 = 0.875
Decrease in price = (1-0.875) x 100 = 12.5%
47.Ramesh started a software business by investing Rs. 50,000. After six months, Rajesh joined him with a capital of Rs. 75,000. After 3 years, they earned a profit of Rs. 27,000. What was Rajesh's share in the profit?
A) 10,000 B) 15,000
C) 20,000 D) 25,000
Answer & ExplanationAnswer: B) 15,000
Explanation:
Ramesh : Rajesh =(50,000*36) : (75,000*30) =4 : 5.
Therefore, Rajesh's share=Rs.(27,000*5/9)= Rs.15,000.
48.In a College there are 1800 students. Last day except 4% of the boys all the students were present in the college. Today except 5% of the girls all the students are present in the college, but in both the day's number of students present in the college, were same. The number of girls in the college is?
A 1000
B. 400
C. 800
D. 600
E. 1200
Aswer and Explanation
Option C
explanation
1 girl + 0.96 boy = 1 boy + 0.95 girl
(1- o.95) girl = (1 - 0.96) boy
0.05 girl = 0.04 boy
girl/boy= 4/5
let us assume that total girls = 4x and boys = 5x
(4x+5x)=1800
x= 200
number of girls is (4*200) = 800
49. manikanta sai buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
A. Rs. 6876.10
B. Rs. 6999.20
C. Rs. 6654
D. Rs. 7000
Answer: Option A
Explanation:
Rebate = 6% of Rs 6650= o.o6*6650= Rs 399
sales tax = 10% of Rs (6650 -399) = Rs(0.1 * 6251) = Rs 625.10
Final amount = Rs (6251 + 625.10) =Rs 6278.10
50.In a test, Bhashyam answer 70 % C++ questions, 40 % C questions and 60 % Java questions correctly. The test had a total of 75 questions, 10 from C++, 30 from C and 35 from Java.
A minimum of 60 % in aggregate was required to be considered for interview. The geek could not clear the test and was not shortlisted for interview. Find by how much marks did the geek miss the interview call,
given that each question was of 1 mark and there was no negative marking for incorrect answers.
Solution : We are given that the geek could answer 70 % C++ questions, 40 % C questions and 60 % Java questions correctly and there were total 75 questions : 10 from C++, 30 from C and 35 from Java.
=> C++ questions answered correctly = 70 % of 10 = 7
=> C questions answered correctly = 40 % of 30 = 12
=> Java questions answered correctly = 60 % of 35 = 21
=> Total questions answered correctly = 7 + 12 + 21 = 40
=> Marks secured = 40 x 1 = 40
Now, marks required = 60 % of 75 = 45
=> Shortfall in marks = 45 – 40 = 5
Therefore, the geek missed the interview call by 5 marks.
1. Raman's salary was decreased by 50% and subsequently increased by 50%. How much percent does he loss.
A. 75
B. 65
C. 45
D. 25
Answer And Explanation
Answer: Option D
Explanation:
Let the origianl salary = Rs. 100
It will be 150% of (50% of 100)
= (150/100) * (50/100) * 100 = 75
So New salary is 75, It means his loss is 25%
2. If x% of y is 100 and y% of z is 200, then find the relation between x and z.
A. z = x
B. 2z = x
C. z = 2x
D. None of above
Answer And Explanation
Answer: Option C
Explanation:
It is , y% of z = 2(x% of y)
=> yz/100 = 2xy/100
=> z = 2x
3. Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first
A. 8%
B. 9%
C. 10%
D. 11%
Answer And Explanation
Answer: Option C
Explanation:
Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10
Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what percent of first number
=> (7x/100 * 10/7x * 100 )% = 10%
4. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate got
A. 55%
B. 56%
C. 57%
D. 58%
Answer And Explanation
Answer: Option C
Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
5. A reduction of 21% in the price of an item enables a person to buy 3 kg more for 100. Thereduced price of item per kg is:
(a) Rs. 5.50
(b) Rs. 7.50
(c) Rs. 10.50
(d) Rs. 7.00
Solution:(d)
Reduced price will be:
Rp/100y per kg
In our case R= Rs. 100 , x=21% , y=3kg
{(100 x 21)/(100 x 3)} = Rs. 7
6. A vessel has 60 L of solution of acid and water having 80% acid. How much water is to be added to make it solution in which acid forms 60%?
(a) 48 L
(b) 20 L
(c) 36 L
(d) None of these
Solution: (b)
Given, percentage of acid = 80%
Then, percentage of water = 20%
In 60L of solution, water = (60 x 20)/ 100 = 12L
Let p liter of water is to be added.
According to the question,
=>{(12 + p)/(60 + p)} x 100 = 40 (∵ 100 – 60 = 40% water)
=>1200 + 100p = 2400 + 40p
⇒ 60p = 1200
p= 20L
7. If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction become 5/2. The original fraction is:
(a) 24/19
(b) 3/18
(c) 95/48
(d) 48/95
Solution: (c)
Let original fraction be p/y
According to the question,
{(120/100)p/(95/100)y} = 5/2
120p/95y = 5/2
=> p/y = (5/2)x (95/120) = 95/48.
8. The monthly income of a person was Rs 13500 and his monthly expenditure was Rs 9000. Nextyear his income increased by 14% and his expenditure increased by 7%. The per cent increase inhis savings was:
(a) 7%
(b) 21%
(c) 28%
(d) 35%
Solution: (c)
Given, monthly income = 13500 and expenditure = 9000
Then, original savings= Rs. (13500-9000) = Rs 4500
New income = 114% of Rs. 13500 = Rs 15390
New expenditure= 107% of Rs 9000 = Rs 9630
New saving = Rs. (15390 – 9630) = Rs 5760
NS = new savings
OS = Original savings
Percentage increase in savings = {(NS – OS)/OS} X 100
{(5760 – 4500)/4500} X 100 = (1260/4500)X 100 = 28%
9. In a quarterly examination a student secured 30% marks and failed by 12 marks. In the same examination another student secured 40% marks and got 28 marks more than minimum marks to pass. The maximum marks in the examination is:
(a) 300
(b) 500
(c) 700
(d) 400
Solution : (d)
here a=12, b=28, x=30%, y= 40%
max marks = 100(a+b)/(x-y)
4000/10 =400
10. In an examination, 52% of the candidates failed in English and 42% failed in Mathematics. If17% failed in both the subjects, then the percentage of candidates who passed in both the subjects, was:
(a) 23%
(b) 21%
(c) 25%
(d) 22%
Solution: (a)
In an examination a% of total number of candidates failed in a subject X and b% of total number of
candidates failed in subject Y and c% failed in both subjects.
Then, percentage of candidates who passed in both the subjects: [100–(a + b– c)%].
Why so?
We need to subtract failed candidate from total to calculate pass candidates, but in this process, we subtract people who were fail in both of subjects twice, so we added c again.
Now equation can be read as [ 100—a-b+c]
Here, a = 52, b = 42 and c = 17, then Percentage of candidates who passed in both the subjects
= [100 –(52 +42–17)]
= 100–77=23%
11. lf A has 4/5 of the number of books that shelf B has. If 25% of the books A are transferred to B and then 25 % of the books from B are transferred to A then the percentage of the total number of books that on shelf A is:
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Solution: (b)
Let the number of books in shelf B be 100.
∴ Number of books in shelf A = {(100 x 4)/5} = 80
On transferring 25% i.e.,of books of shelf A to shelf B, the books on
shelf B = {100 + (80 x 25)/100}
B = 100 + 20 = 120
Books left in shelf A = 80-20 = 60
Again, on transferring 1/4th books of shelf B to shelf A, the books on
shelf A = {60 + (120/4)} = 90
Total no of books in A and B = 120 +60 = 180
Required percentage of books on shelf A = (90/180) x 100 = 50%
12. Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calcualte average marks of the remaining students of the class.
A. 60
B. 70
C. 75
D. 80
Answer And Explanation
Answer: Option C
Explanation:
Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
=> 70x = 8000 - (950 + 1800) = 5250
=> x = 75.
13. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are
A. 42,30
B. 42,31
C. 42,32
D. 42,33
Answer And Explanation
Answer: Option D
Explanation:
Let their marks be (x+9) and x.
Then, x+9 = 56/100(x + 9 +x)
=> 25(x+9)
=> 14 (2x + 9)
=> 3x = 99
=> x = 33.
So, their marks are 42 and 33
14. One fourth of one third of two fifth of a number is 15. What will be40% of that number
A. 140
B. 150
C. 180
D. 200
Answer And Explanation
Answer: Option C
Explanation:
(1/4) * (1/3) * (2/5) * x = 15 then x = 15 * 30 = 450
40% of 450 = 180
15. 10% of inhabitants of a village having died of cholera, a panic set in, during which 25% of the remaining inhabitants let the village. The population is then reduced to 4050. Find the original inhabitants
A. 5500
B. 6000
C. 6500
D. 7000
Answer And Explanation
Answer: Option B
Explanation:
Let the total number is x,
then,
(100-25)% of (100 - 10)% x = 4050
=> 75% of 90% of x = 4050
=> 75/100 * 90/100 * x = 4050
=> x = (4050*50)/27 = 6000
16. What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A.
1
B.
14
C.
20
D.
21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = 14 x 100 % = 20%.
70
17.
What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
A.
1
B.
14
C.
20
D.
21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = 14 x 100 % = 20%.
70
18.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
A.
2700
B.
2900
C.
3000
D.
3100
Answer: Option A
Explanation:
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (0.45 x 6000) = 2700.
19.
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A.
Rs. 200
B.
Rs. 250
C.
Rs. 300
D.
None of these
Answer: Option B
Explanation:
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
z + 1.20z = 550
11/5z = 550
z = 550/11 x 5 = 250.
20.
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
A.
4.37%
B.
5%
C.
6%
D.
8.75%
Answer: Option B
Explanation:
Increase in 10 years = (262500 - 175000) = 87500.
Increase% = 87500/175000 x 100 % = 50%.
Required average = 50/10 % = 5%.
21.
Two persons Raj and Ramu started working for a company in similar jobs on January 1, 1991. Raj's initial monthly salary was Rs 400, which increases by Rs 40 after every year. Ramu's initial monthly salary was Rs 500 which increases by Rs 20 after every six months. If these arrangements continue till December 31, 200. Find the total salary they received during that period.
A. Rs 1,08,000
B. Rs 1,44,000
C. Rs 1,32,000
D. Rs 1,52,400
Solution:
Option(D) is correct
Raj's salary as on 1 jan 1991 is Rs 400 per month
His increment in his month salary is Rs 40 per annum
His total salary from 1 jan 1991 to 31st dec 2000
i.e. in ten years
=12[2(400)+(10−1)40]×102
=Rs 69,600
Ramu's salary as on Jan 1st 1991 is Rs 550 and his half yearly increment in his month salary is Rs 20.
His total salary from 1 jan 1991 to dec 31, 2000
=6[2(500)+(20−1)20]×202
=Rs 82,000
Total salary of Raj and Ramu in the ten year period:
= Rs. 69600+ Rs. 82800
⇒ Rs 1,52,400
22.
30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 15%
B. 20%
C. 80%
D. 70%
Solution:
Option(C) is correct
20% of the men are above the age of 50 years. 20% of these men play football. Therefore, 20% of 20% or 4% of the total men are football players above the age of 50 years.
20% of the men are football players. Therefore, 16% of the men are football players below the age of 50 years.
Therefore, the % of men who are football players and below the age of
50=16/20×100
23.
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6, then what was the cost of the tax free items?
A. Rs. 15
B. Rs. 15.70
C. Rs. 19.70
D. Rs. 20
Solution:
Option(C) is correct
Let the amount taxable purchases be Rs. x
Then,
6% of x=30100
x=30100×1006
=5
Cost of tax free items
= Rs. [25−(5+0.30)]
= Rs. 19.70
24.
If the cost price of 20 articles is equal to the selling price of 16 articles, What is the percentage of profit or loss that the merchant makes?
A. 20%Profit
B. 25% Loss
C. 25% Profit
D. 33.33% Loss
Solution:
Option(C) is correct
Let Cost price of 1 article be Re.1.
Therefore, Cost price of 20 articles = Rs. 20.
Selling price of 16 articles = Rs. 20
Therefore, Selling price of 20 articles
=2016×20
=25
Profit = Selling price - Cost price
=25−20=5
Percentage of profit=profitC.P×100
=520×100=520×100
= 25% Profit
25.
A man purchased a bag for Rs.360 and sold it the same day for Rs. 360, allowing the buyer a credit of 9 years.
If the interest be 712% then what is the man's gain?
A. 159
B. 173
C. 243
D. 287
Solution:
Option(C) is correct
Interest rate =152=7.5%
CP=Rs. 360.
SP=360+intrest on 360 for 9 years
Interest = 360×15×92×100
=Rs. 243.
Gain = Interest =Rs. 243.
26.
Two merchants sell, each an article for Rs.1000. If Merchant AA computes his profit on cost price, while Merchant BB computes his profit on selling price, they end up making profits of 25% respectively.
By how much is the profit made by Merchant BB greater than that of Merchant AA?
A. Rs.50.00
B. Rs.66.67
C. Rs.75.00
D. Rs.150.00
Solution:
Option(A) is correct
Merchant B computes his profit as a percentage of selling price. He makes a profit of 25% on selling price of Rs.1000. i.e. his profit =25/1000
Merchant A computes his profit as a percentage of cost price.
If he makes a profit of 25% or 1/4th of his cost price, then his profit expressed as a percentage of selling price,
=11+4=(15)th or 20% of selling price.
So, Merchant A makes a profit of 20% of Rs. 1000=Rs. 200
Hence, Merchant B makes Rs.50Rs.50 more profit than Merchant A.
27.
After offering a discount of 37.5, Pankaj sold the rice at a profit of 25. Had he offered a discount of 41.67, his profit or loss percent would have been:
A. 16.66% profit
B. 12% loss
C. 29.17% loss
D. 8.33% profit
Solution:
Option(A) is correct
Let the marked price of the rice = 8p
Discount
=37.5%=38×MP
Selling price
=8p−38×8p=5p
5p=CP+25100×25
⇒CP=4p
If he had offered a discount of 41.67%=512
SP=8p−512×8p
=14p3
Profit
=14p3−4p
=2p3
Profit percentage
=2p/34p×100
=1006%
=16.66%
28.
The cost of raw material of a product increases by 30, the manufacturing cost increases by 40 and the selling price of the product increases by 60. The raw material and the manufacturing cost, originally, formed 40 and 60 of the total cost respectively. If the original profit was one-fourth the original manufacturing cost, find the approximation new profit percentage.
A. 48.39%
B. 54.68%
C. 62.48%
D. 42.36%
Solution:
Option(A) is correct
Let the total initial cost of the product be Rs 100.
Manufacturing cost = Rs 60
Raw materials cost = Rs 40
Also, original selling price
= Rs100+25% of 60=Rs 115
New raw material cost
= Rs 40+30% of 40= Rs 52
New manufacturing cost
== Rs 60+20% of 60= Rs 72
New cost of the product = Rs124
New selling price
=115+60% of 115=Rs 184
New profit percentage
=60124×100
=48.38%
29.
In a period from January to March, Jamshedpur Electronics sold 3150 units of Television, having started with a beginning inventory of 2520 units and ending with an inventory of 2880. What was the value of order placed (Rupees in thousands) by Jamshedpur Electronics during the three months period? [Profits are 2525 of cost price, uniformly.]
A. 26325
B. 22320
C. 25200
D. 28080
Solution:
Option(D) is correct
Units ordered = Units sold + Ending Inventory-Beginning Inventory
⇒3150+2880−2520=3510
Total sales of Television in Rs. Thousand
⇒900+1800+6300+1050+2100+7350+1200+2400+8400=31500
Sales Price per unit of Television in Rs. Thousand
= 31500/3150=10
Profits are 25%of the cost price.
Sales Price = Cost Price + Profits = Cost Price +0.25×+0.25× Cost Price =1.25×=1.25× Cost Price
Cost Price per unit of Television
=Sales Price per unit1.25
=101.25
=8
The value of the order placed in Rs. Thousand = Units ordered ×× Cost Price per unit
⇒3510×8=28080
30.
A Techno company has 14 machines of equal efficiency in its factory. The annual manufacturing costs are Rs 42,000 and establishment charges are Rs 12,000. The annual output of the company is Rs 70,000.
The annual output and manufacturing costs are directly proportional to the number of machines. The shareholders get 12.5 profit, which is directly proportional to the annual output of the company.
If 7.14 machines remain closed throughout the year, then the percentage decrease in the amount of profit of the shareholders would be:
A. 12%
B. 12.5%
C. 13.5%
D. None of these
Solution:
Option(B) is correct
Original profit
= 70,000−42,000−12,000=16,000
If 7.14% of 14 i.e. one of the machines closed throughout the year, then change in profit will be:
=1314×[70,000−42,000]
=14,000
Thus, decrease in the profit %%
=2000/16000×100
=12.5%
31.
If the price of petrol increases by 25 and Raj intends to spend only an additional 15 on petrol, by how much will he reduce the quantity of petrol purchased?
A. 10%
B. 12%
C. 8%
D. 6.67%
Solution:
Option(C) is correct
Let the price of 1 litre of petrol be Rs. x and let Raj initially buys 'y litres of petrol.
Therefore, he would have spent Rs. xy on petrol.
When the price of petrol increases by 25%, the new price per litre of petrol is 1.25x.
Raj intends to increase the amount he spends on petrol by 15%
i.e., he is willing to spend xy+15% of xy=1.15xy
Let the new quantity of petrol that he can get be 'q.
Then, 1.25x×q=1.15xy
q=1.15xy1.25x
=1.15y1.25
=0.92y
As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what he used to get earlier.
Or a reduction of 8%
32.
A shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by xx during the year. A famine hits his village in the next year and many of his sheep die. The sheep population decreases by yy during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheep. Which of the following is correct?
A. x>y
B. y>x
C. x=y
D. Cannot be determined
Solution:
Option(A) is correct
Let us assume the value of x to be 10%
Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million
In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.
i.e., 0.1 million sheep have died in 2001.
In terms of the percentage of the number of sheep alive at the beginning of 2001,
it will be (0.1/1.1)×100%=9.09%(0.1/1.1)×100%=9.09%.
From the above illustration it is clear that x>y
33.
Peter got 30 of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40 of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
A. 35
B. 250
C. 75
D. 85
Solution:
Option(D) is correct
Let x be the maximum marks in the examination.
Therefore, Peter got 30% of x
=30100×x=0.3x
And Paul got 40% of x
=40100×x
=0.4x
In terms of the maximum marks Paul got 0.4x−0.3x=0.1x more than Peter. --------(1)
The problem however, states that Paul got 15 marks more than the passing mark and Peter got 10 marks less than the passing mark.
Therefore, Paul has got 15 + 10 = 25 marks more than Peter.-------- (2)
Equating (1) and (2), we get
0.1x=25
⇒x=250
xx is the maximum mark and is equal to 250 marks.
We know that Peter got 30% of the maximum marks.
Therefore, Peter got
=30100×250
=75 marks
We also know that Peter got 10 marks less than the passing mark.
Therefore, the passing mark will be 10 marks more than what Peter got = 75+10=85
34.
In an acoustics class, 120 students are male and 100 students are female. 25% of the male students and 20% of the female students are engineering students.
20% of the male engineering students and 25% of the female engineering students passed the final exam.
What percentage of engineering students passed the exam?
A. 5%
B. 10%
C. 16%
D. 22%
Solution:
Option(D) is correct
There are 100 female students in the class, and 20% of them are Engineering students.
Now, 20% of 100 equals (20100)×100=20..
Hence, the number of female engineering students in the class is 20.
Now, 25% of the female engineering students passed the final exam:
25.
Hence, the number of female engineering students who passed is 5.
There are 120 male students in the class. And 25% of them are engineering students.
Now, 25% of 120 equals(25100)×120=(14)×120=30
Hence, the number of male engineering students is 30.
Now, 20% of the male engineering students passed the final exam:
20
Hence, the number of male engineering students who passed is 6.
Hence, the total number of engineering students who passed is:the total number of engineering students who passed is:
(Female Engineering students who passed)+(Female Engineering students who passed)+
(Male Engineering students who passed)(Male Engineering students who passed)
=5+6=11
The total number of engineering students in the class is:The total number of engineering students in the class is:
(Number of female engineering students)+(Number of female engineering students)+
(Number of male engineering students)=(Number of male engineering students)=
=30+20=50
Hence, the percentage of engineering students who passed is
(Total number of engineering students who passedTotal number of engineering students)×100(Total number of engineering students who passedTotal number of engineering students)×100
=(1150)×100
= 22%
35.
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 45%
B. 45 5/11%
C. 54 6/11%
D. 55%
Solution:
Option(B) is correct
Number of runs made by running,
=110−(3×4+8×6)=110−(3×4+8×6)
=110−(60)=110−(60)
=50.=50.
Required percentage,
=(50110×100)%=(50110×100)%
=45 5/11%
36.
If bb equals 10% of a and c equals 20% of b, then which one of the following equals 30% of c?
A. 0.006% of a
B. 0.006% of a
C. 0.06% of a
D. 0.6% of a
Solution:
Option(D) is correct
b=10% of a=(10100)×a=0.1aa=(10100)×a=0.1a
c =20% of b= =(20/100)×b
=0.2b=0.2×0.1a=0.2b=0.2×0.1a
Now, 30% of c=(30/100)×c
=0.3c=(0.3)(0.2)(0.1a)
=0.6%a
37.
8 is 4% of a, and 4 is 8% of b. c equals b/a. What is the value of c?
A. 1/32
B. ¼
C. 1
D. 4
Solution:
Option(B) is correct
4% of a is 4a/100.
Since this equals 8, we have 4a/100=8.
Solving for a yields a=8×(100/4)=200.
Also, 8% of bb equals 8b/100 and this equals 4.
Hence, we have (8/100)×b=4.
Solving for b yields b=50
Now, c=ba
=50/200
=1/4.
38.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid.
If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
A. 2700
B. 2900
C. 3000
D. 3100
Solution:
Option(A) is correct
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
=(45/100)×6000 = 2700.
39.
If 50% of xx equals the sum of yy and 20, then what is the value of x–2yx–2y?
A. 20
B. 40
C. 60
D. 80
Solution:
Option(B) is correct
50% of xx equals the sum of yy and 20. Expressing this as an equation yields:
(50/100)×x=y+20
x/2=y+20
x=2y+40
x–2y= 40
40.
The price of an item changed from Rs. 120 to Rs. 100. Then later the price decreased again from Rs. 100 to Rs. 80.
Which can be said about the two decreases in percentage term?
A. 1st decrease is larger
B. 2nd decrease is larger
C. both are equal
D. nothing can be said
Solution:
Option(B) is correct
First decrease in percent,
partwhole=(120−100)120
=0.17=17%
Second decrease in percent,
partwhole=(100−80)100
=0.20=20%
The second decrease is larger in percent term.
The parts were the same in both cases, but the whole was smaller in the second decrease.
41.
Sahil earns 10 percent more than Satish and satish earns 20 percent more than Swati. if swati earns rs 17,500 less than Sahil, what is the earnings of each?
A. Swati = Rs. 35,900.0, Sahil = Rs. 53,400.0, Satish = Rs. 58,740.0
B. Swati = Rs. 39,500.0, Sahil = Rs. 57,000.0, Satish = Rs. 62,700.0
C. Swati = Rs. 54,687.5, Sahil = Rs. 72,187.5, Satish = Rs. 65,625.0
D. Swati = Rs. 69125.0, Sahil = Rs. 86,625.0, Satish = Rs. 95,287.5
Solution:
Option(C) is correct
Let Swati earns Rs. x.
Satish earns =x+20/100
Sahil earns =1.2x+10/100
Earning difference of Sahil and Swati =17,500
1.32x−x=17,500
0.32x=17,500
⇒x=175000.32
⇒x=Rs. 54,687.5
So, Swati's earnings,
x=Rs. 54,687.5
Sahil's earnings,
=1.32×x
=1.32×Rs. 54,687.5
=Rs. 72,187.5
Satish's earnings,
=1.2×x
=1.2×Rs. 54,687.5
=Rs. 65,625.0
42.
A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest.
What percent of his apples does the vendor throw?
A. 17
B. 23
C. 77
D. None of these
Solution:
Option(B) is correct
Let the number of apples be 100.
On the first day he sells 60% apples i.e.,60 apples. Remaining apples =40.
He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has 40−6=3440−6=34 apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in total he throws 6+17=6+17= 23 apples.
43.Fresh grapes contain 60% water, while dry grapes contain 40% water. How many kgs of dry grapes can be obtained by 60 kg of fresh grapes?
A) 25 kgs B) 30 kgs
C) 35 kgs D) 40 kgs
Answer & ExplanationAnswer: D) 40 kgs
Explanation:
The fruit content in both the fresh fruit and dry fruit is the same.
Let 'p' kgs of dry grapes are obtained from the 60 kgs of fresh grapes
=> 60 x 40/100 = p x 60/100=> p = 40 kgs.
Therefore, 40 kgs of Dry grapes are obtained from 60 kgs of Fresh grapes.
44.In Veeru Bhai Pvt.limited company 60% of the employees are men and 48% of the employees are Engineer and 66.6% of these are men. The percentage of women who are not engineers:
A) 33.33% B) 60%
C) 52% D) 46.66%
Answer & ExplanationAnswer: B) 60%
Explanation:
Men Women
600x 400x
total engineers = 480x
Male engineers = 480x X 0.66 = 320
Women who are Engineers = 160x
Women are not Engineers = 400x - 160x = 240x
Required percentage = 240/400x100 = 60%
45.A man lost half of its initial amount in the gambling after playing 3 rounds. The rule of gambling is that if he wins he will receive Rs. 100, but he has to give 50% of the total amount after each round.Luckily he won all the three rounds. The initial amount with which he had started the gambling was :
A) 500/3 B) 700/3
C) 300 D) 600
Answer & ExplanationAnswer: B) 700/3
Explanation:
Let the initial amount be x (with gambler), then
[[(x+100)1/2 + 100]12+100]1/2 = x/2
=> x = 700/3
46.The Shopkeeper increased the price of a product by 25% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 70% of the required amount. What is the net difference in the expenditure on that product?
A) 10% more B) 5% more
C) 12.5% more D) 17.5% more
Answer & ExplanationAnswer: C) 12.5% more
Explanation:
Quantity X Rate = Price
1 x 1 = 1
0.7 x 1.25 = 0.875
Decrease in price = (1-0.875) x 100 = 12.5%
47.Ramesh started a software business by investing Rs. 50,000. After six months, Rajesh joined him with a capital of Rs. 75,000. After 3 years, they earned a profit of Rs. 27,000. What was Rajesh's share in the profit?
A) 10,000 B) 15,000
C) 20,000 D) 25,000
Answer & ExplanationAnswer: B) 15,000
Explanation:
Ramesh : Rajesh =(50,000*36) : (75,000*30) =4 : 5.
Therefore, Rajesh's share=Rs.(27,000*5/9)= Rs.15,000.
48.In a College there are 1800 students. Last day except 4% of the boys all the students were present in the college. Today except 5% of the girls all the students are present in the college, but in both the day's number of students present in the college, were same. The number of girls in the college is?
A 1000
B. 400
C. 800
D. 600
E. 1200
Aswer and Explanation
Option C
explanation
1 girl + 0.96 boy = 1 boy + 0.95 girl
(1- o.95) girl = (1 - 0.96) boy
0.05 girl = 0.04 boy
girl/boy= 4/5
let us assume that total girls = 4x and boys = 5x
(4x+5x)=1800
x= 200
number of girls is (4*200) = 800
49. manikanta sai buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
A. Rs. 6876.10
B. Rs. 6999.20
C. Rs. 6654
D. Rs. 7000
Answer: Option A
Explanation:
Rebate = 6% of Rs 6650= o.o6*6650= Rs 399
sales tax = 10% of Rs (6650 -399) = Rs(0.1 * 6251) = Rs 625.10
Final amount = Rs (6251 + 625.10) =Rs 6278.10
50.In a test, Bhashyam answer 70 % C++ questions, 40 % C questions and 60 % Java questions correctly. The test had a total of 75 questions, 10 from C++, 30 from C and 35 from Java.
A minimum of 60 % in aggregate was required to be considered for interview. The geek could not clear the test and was not shortlisted for interview. Find by how much marks did the geek miss the interview call,
given that each question was of 1 mark and there was no negative marking for incorrect answers.
Solution : We are given that the geek could answer 70 % C++ questions, 40 % C questions and 60 % Java questions correctly and there were total 75 questions : 10 from C++, 30 from C and 35 from Java.
=> C++ questions answered correctly = 70 % of 10 = 7
=> C questions answered correctly = 40 % of 30 = 12
=> Java questions answered correctly = 60 % of 35 = 21
=> Total questions answered correctly = 7 + 12 + 21 = 40
=> Marks secured = 40 x 1 = 40
Now, marks required = 60 % of 75 = 45
=> Shortfall in marks = 45 – 40 = 5
Therefore, the geek missed the interview call by 5 marks.
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