# RRB JE Percentages Questions PDF

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## RRB JE Percentages Questions PDF

Download Top 20 RRB JE Percentage Questions and Answers PDF. RRB JE Percentage questions based on asked questions in previous exam papers very important for the Railway JE exam

Question 1: What is the value of 8.33% of 72.72% of 28.57% of 462 ?

a) 2

b) 4

c) 6

d) 8

Question 2: What is the value of 33.33% of 44.44% of 37.5% of 216 ?

a) 12

b) 15

c) 18

d) 21

Question 3: Average of goals scored by Marcelo in 8 laliga matches is 3.75. If in his 9th,10th and 11th matches he scores 1,2 an 3 goals respectively . What is the percentage change in his average after 11 matches?

a) 12.2%

b) 12.8%

c) 12.4%

d) 12.8%

Question 4: Cost of an article is decreased by 44.44 %,increased by 66.66% and finally increased by 8.33%. What is the percentage change in the cost of the article ?

a) 0.01%

b) 0.1%

c) 0.3%

d) 0.2%

Question 5: Length and breadth of a rectangle is increased by 10% and 20% respectively. What is the percentage change in the area ?

a) 30%

b) 31%

c) 32%

d) 33%

Question 6: Find 80% of 40% of 150% of 2000 ?

a) 920

b) 940

c) 960

d) 980

Question 7: 257% of 489 = 489% of X, then find X ?

a) 128.5

b) 257

c) 385.5

d) 514

Question 8: If the cost price of 12 oranges is equal to selling price of 10 oranges, then the percentage of profit is

a) $16\frac{2}{3}$%

b) $18$%

c) $20$%

d) $25$%

Question 9: A basket has only apples and oranges. 30% of the fruits are apples. 60% of apples and 70% of oranges are ripe. What percentage of the total fruits is not ripe?

a) 33

b) 67

c) 35

d) 65

Question 10: 30% of a class consists of girls. If 20% of the girls and 30% of the boys in that class are selected to form a team, what is the percentage of girls in the team?

a) 33.33%

b) 22.22%

c) 19.19%

d) 15.15%

Question 11: The cost price of 18 articles is equal to the selling price of 15 articles. The gain percent is–

a) 15%

b) 20%

c) 25%

d) 18%

Question 12: The respective ratio between the present ages of two persons A and G is 3 : 4. After 14 years the age of G is 70 years. What will be the present age of A?

a) 38 years

b) 40 years

c) 42 years

d) 56 years

e) None of these

Question 13: Pranita got 30 marks more in Math than what she got in Science. Her Math marks are 60% of the sum of her Math and Science marks. What are her Science marks?

a) 90

b) 150

c) 120

d) 60

Question 14: In a class of 50 students there are 27 boys. The average weight of these boys is 72 Kg and average weight of the full class is 55.44 kgs. What is the average weight (in kgs) of the girls of the class?

a) 42

b) 48

c) 30

d) 36

Question 15: Due to increase of k% in the side, the area of a square increases by 69%. What is the value of k?

a) 30

b) 33

c) 34.5

d) 35

Question 16: How much water (in litres) must be added to 80 litres solution of milk and water containing 10% milk, so that it becomes a 5% milk solution?

a) 10

b) 20

c) 40

d) 80

Question 17: What is the discount percentage offered on a book having marked price Rs 2150 being sold at Rs 1892?

a) 12

b) 13

c) 14

d) 16

Question 18: 40 % of 80 % of 50 % of 800 is

a) 112

b) 164

c) 138

d) 128

Question 19: A, B and C are three students. A got 18% more marks than B and 12% less than C. If B got 220 marks, then how much marks C has got?

a) 230

b) 295

c) 240

d) 290

Question 20: The number of trees in a town is 17640. If the numbers of trees increases annually at the rate of 5%, then how many trees were there 2 years ago?

a) 14000

b) 15000

c) 16000

d) 19450

8.33%=1/12
72.72%=8/11
28.57%=2/7
Therefore we have (1/12)*(8/11)*(2/7)*462= (1/12)(8/11)(2/7)*11*7*6
=8

33.33%=1/3
44.44%=4/9
37.5%=3/8
Therefore we have (1/3)(4/9)(3/8)*216=
=12

His old average=3.75
Total goals=3.75*8
=30 goals
His new average is (30+1+2+3)/(11)=3.27
Percentage change in his average=(3.75-3.27)*100/3.75
=0.128*100
=12.8%
Hence, option D is the correct answer.

let the cost be x
44.44%=4/9
66.66%=2/3
8.33%=1/12
So the change value will be x(1-(4/9))(1+(2/3))(1+(1/12))
=x*(5/9)(5/3)(13/12)
=1.003x
Percentage change (1.003x-x)*100=0.3%
Hence, option C is the correct answer.

let us assume length and breadth to be l and b.
Area=l*b
new length =1.1l
new area=1.1l*1.2b
=1.32lb
Percentage change=((1.32lb-lb)/lb)*100
=32%

80% of 40% of 150% of 2000 = (0.8)(0.4)(1.5)(2000) = (0.32)(3000) = 960

So the answer is option C.

x % of y = y % of x

similarly, 257% of 489 = 489% of 257

So the answer is option B.

Let the cost of each orange be y.
Cost of 10 apples =10y
Selling price of 10 apples =Cost of 12 apples =12y
The profit for selling 10 apples=12y-10y=2y
Thus profit%=(2yx100)/10y= 20%

Suppose, there are 100 fruits in the basket.
=> 30 are apples and 70 are oranges
60% of apples are ripe => 30 * 0.6 = 18 apples are ripe => 12 apples are not ripe.
70% of oranges are ripe => 70 * 0.7 = 49 oranges are ripe => 21 oranges are not ripe.
=> Total number of fruits that are not ripe = 12 + 21 = 33
=> 33% of fruits are not ripe

Let the number of students in the class be 100.
Number of girls = 30
Number of boys = 70
Number of girls selected to the team = 20% of 30 = 6
Number of boys selected to the team = 30% of 70 = 21
So, percentage of girls in the team = 6/27 * 100% = 22.22%

Let the price be 180 Rs. 18 items were purchased at 180 rupees and 15 items were sold at 180 rupees.

So, cost price is 10 rupees and selling price is 12 rupees.

So, profit percent = 20%

Let she got $x$ marks in Science

=> Maths marks = $(x+30)$

According to ques,

=> $(x+30)=\frac{60}{100}(x+x+30)$

=> $(x+30)=\frac{3}{5}(2x+30)$

=> $5x+150=6x+90$

=> $6x-5x=150-90$

=> $x=60$

=> Ans – (D)

Total number of students = 50 and number of boys = 27

=> Number of girls in class = 50 – 27 = 23

Average weight of boys = 72 kg

=> Total weight of boys = $72 \times 27 = 1944$ kg

Similarly, total weight of full class = $55.44 \times 50 = 2772$ kg

=> Total weight of girls = 2772 – 1944 = 828 kg

$\therefore$ Average weight of girls = $\frac{828}{23} = 36$ kg

=> Ans – (D)

Let side of square = $10$ cm

=> Area of square = $(10)^2=100$ $cm^2$

Area is increased b 69%, => New area = $100+(\frac{69}{100}\times100)=169$ $cm^2$

Thus, new side of square = $\sqrt{169}=13$ cm

$\therefore$ Increase in side = $k=\frac{(13-10)}{10}\times100$

= $30\%$

=> Ans – (A)

Quantity of mixture = 80 litres

Quantity of milk = $\frac{10}{100}\times80=8$ litres

Let $x$ litres of water is added

According to ques,

=> $\frac{5}{100}(x+80)=8$

=> $x+80=8\times20=160$

=> $x=160-80=80$ litres

=> Ans – (D)

Marked price = Rs. 2150

Selling price = Rs. 1892

=> Discount % = $\frac{(2150-1892)}{2150}\times100$

= $\frac{516}{43}=12\%$

=> Ans – (A)

50 % of 800 = 400
80 % of 400 = 320
40 % of 320 = 32*40/100 = 128
Hence, option D is the correct answer.

B’s marks = 220

A got 18% more marks than B

=> A’s marks = $220+(\frac{18}{100}\times220)=259.6$

Also, A got 12% less marks than C

=> C’s marks = $259.6\times\frac{100}{(100-12)}$

= $\frac{25960}{88}=295$

=> Ans – (B)

Let the number of trees 2 years ago = $P$

Rate of increase of trees = 5%

=> Trees after 2 years = $P(1+\frac{R}{100})^T$

=> $P(1+\frac{5}{100})^2=17640$

=> $P(\frac{21}{20})^2=17640$

=> $P=17640\times\frac{400}{441}$

=> $P=40\times400=16000$

=> Ans – (C)

We hope this Percentage Questions for RRB JE Exam will be highly useful for your preparation.