# TISSNET Percentage Questions [Download PDF]

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**Question 1: **A group of 20 girls has average age of 12 years. Average of first 12 from the same group is 13 years and what is the average age of other 8 girls in the group?

a) 10

b) 11

c) 11.5

d) Cannot be determined

e) None of these

**1) Answer (E)**

**Solution:**

Let the age of each of the girl in the group be $x_1,x_2,x_3,…..,x_{20}$ years

Average age of 20 girls = 12

=> $\frac{(x_1+x_2+x_3+…..+x_{20})}{20}=12$

=> $(x_1+x_2+x_3+…..+x_{20})=12 \times 20=240$ ————(i)

Average of first 12 girls = 13

=> $\frac{(x_1+x_2+x_2+….+x_{12})}{12}=13$

=> $(x_1+x_2+x_3+…..+x_{12})=13 \times 12=156$ ———–(ii)

Subtracting equation (ii) from (i)

=> $(x_1+x_2+x_3+…..+x_{20})$ $-$ $(x_1+x_2+x_3+…..+x_{12}) = (240-156)$

=> $(x_{13}+x_{14}+…..+x_{20})=84$

Dividing above equation by 8

=> $\frac{(x_{13}+x_{14}+……+x_{20})}{8}=\frac{84}{8} = 10.5$

=> Ans – (E)

**Question 2: **The total marks obtained by Jaya in the maths and Physics together are 50 more than the marks obtained by her in chemistry, if she got 20 marks in the physics .What are her marks in maths?

a) 20

b) 40

c) 80

d) Cannot Determined

e) None of these

**2) Answer (D)**

**Solution:**

Let marks obtained by Jaya in Physics, maths and chemistry respectively be $p,m,c$

Marks in physics, $p=20$

According to ques, => $m+p=c+50$

=> $m+20=c+50$

=> $m-c=50-20=30$

There is only 1 equation and 2 variables, thus we cannot determine the marks scored by her in maths or chemistry.

=> Ans – (D)

**Question 3: **Two chairs and three tables cost Rs 1025/- and 3 chairs and 2 tables cost is Rs 1,100/- .What is the difference between the cost one table and chair?

a) 75

b) 35

c) 125

d) Cannot be determined

e) None of these

**3) Answer (A)**

**Solution:**

Let cost price of 1 chair = $Rs.$ $x$ and cost price of 1 table = $Rs.$ $y$

=> $2x+3y=1025$ ———–(i)

and $3x+2y=1100$ ——–(ii)

Subtracting equation (i) from (ii)

=> $(3x-2x)+(2y-3y)=(1100-1025)$

=> $x-y=75$

$\therefore$ Difference between the cost one table and chair = Rs. 75

=> Ans – (A)

**Question 4: **The cost of 4 cell phones and 7 digital cameras is Rs 1,25,627.what is the cost of 8 cellphones and 14 digital cameras.

a) 251254

b) 252627

c) 225524

d) Cannot determined

e) None of these

**4) Answer (A)**

**Solution:**

Let cost of 1 cell phone = $Rs. x$ and 1 digital camera = $Rs. y$

=> $4x + 7y = 1,25,627$

Multiplying both sides by 2, we get :

=> $8x + 14y = 125627 \times 2 = 251254$

$\therefore$ Cost of 8 cellphones and 14 digital cameras = Rs. 2,51,254

**Question 5: **If a amount of Rs 41910 is distributed amongst 22 persons equally,how much each person get

a) 1905

b) 2000

c) 1885

d) 2015

e) None of these

**5) Answer (A)**

**Solution:**

Amount distributed = Rs. 41910

Number of person = 22

Each person will get = $\frac{41910}{22}$

= Rs. 1,905

**Question 6: **A canteen requires 112 kgs of wheat for week .how many kgs of wheat requires for 69 days

a) 1204

b) 1401

c) 1104

d) 1014

e) None of these

**6) Answer (C)**

**Solution:**

Wheat requirement for 7 days = 112 kg

=> Wheat requirement for 1 day = $\frac{112}{7} = 16$ kg

$\therefore$ Wheat requirement for 69 days = $16 \times 69$

= 1104 kg

**Question 7: **The difference between 42% of a number and 28% of the same number is 210.what is the 59% of that number

a) 630

b) 885

c) 420

d) 900

e) none of these

**7) Answer (B)**

**Solution:**

Let the number = $100x$

Acc. to ques,

=> $(\frac{42}{100} \times 100x) – (\frac{28}{100} \times 100x) = 210$

=> $42x – 28x = 14x = 210$

=> $x = \frac{210}{14} = 15$

=> Number = $100 \times 15 = 1500$

$\therefore$ 59 % of the number = $\frac{59}{100} \times 1500$

= 885

**Question 8: **The average of the 5 consecutive even numbers A,B,C,D ,E is 52.what is the product of B & E

a) 2912

b) 2688

c) 3024

d) 2800

e) NONE OF THESE

**8) Answer (D)**

**Solution:**

Let the five consecutive even numbers A,B,C,D ,E = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$ respectively.

Average = $\frac{A+B+C+D+E}{5} = 52$

=> $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 52 \times 5$

=> $5x = 52 \times 5$

=> $x = \frac{52 \times 5}{5} = 52$

=> $B = 52 – 2 = 50$ and $E = 52 + 4 = 56$

$\therefore$ Product of B & E = $50 \times 56 = 2800$

**Question 9: **Pinku, Rinku and Tinku divide an amount of 4200 amongst themselves in the ratio of 7:8:6 respectively. If an amount 200 is added to their shares. What will be the new ratio?

a) 8:9:6

b) 7:9:5

c) 7:8:6

d) 8:9:7

e) None of these

**9) Answer (D)**

**Solution:**

Let the amount received by Pinku, Rinku and Tinku respectively = $7x , 8x , 6x$

Total amount = $7x + 8x + 6x = 4200$

=> $21x = 4200$

=> $x = \frac{4200}{21} = 200$

=> Amount with Pinku = 7*200 = 1400

Rinku = 8*200 = 1600

Tinku = 6*200 = 1200

If Rs. 200 is added to their shares, respective ratio

= $(1400 + 200) : (1600 + 200) : (1200 + 200)$

= $1600:1800:1400 = 8:9:7$

**Question 10: **find the average of set scores? 221,231,441,359,665,525

a) 399

b) 428

c) 407

d) 415

e) None of these

**10) Answer (C)**

**Solution:**

Set : 221,231,441,359,665,525

Sum = 221 + 231 + 441 + 359 + 665 + 525 = 2442

=> Required average = $\frac{2442}{6}$

= 407

**Question 11: **45% of a number is 255.6 .what is the 25% of that number?

a) 162

b) 132

c) 152

d) 142

e) None of these

**11) Answer (D)**

**Solution:**

Let number = $100x$

Acc. to ques,

=> $\frac{45}{100} \times 100x = 255.6$

=> $45x = 255.6$

=> $x = \frac{255.6}{45} = 5.68$

Number = $100 \times 5.68 = 568$

$\therefore$ 25% of number = $\frac{25}{100} \times 568$

= 142

**Question 12: **Monthly salaries of Pia and Som are in the respective ratio of 5: 4. Pia, from her monthly salary, gives $\frac{3}{5}$th to her mother. 15% towards her sister’s tuition fees, 18% towards a loan and she shops with the remaining amount which was Rs. 2,100. What is the monthly salary of Som ?

a) Rs.25,000

b) Rs.30.000

c) Rs.15,000

d) Rs.20,000

e) Rs.24,000

**12) Answer (E)**

**Solution:**

Let monthly salary of Pia = $Rs. 500x$

=> Monthly salary of Som = $Rs. 400x$

Amount given by Pia to her mother = $\frac{3}{5} \times 500x = 300x$

Amount for her sister tuition fees = $\frac{15}{100} \times 500x = 75x$

Amount for loan = $\frac{18}{100} \times 500x = 90x$

=> Amount left = $500x – (300x + 75x + 90x) = 2100$

=> $500x – 465x = 35x = 2100$

=> $x = \frac{2100}{35} = 60$

$\therefore$ Som’s salary = $400 \times 60 = Rs. 24,000$

**Question 13: **There are three positive numbers, ${1 \over 3}$rd of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number?

a) 11

b) 14

c) 10

d) 9

e) 13

**13) Answer (A)**

**Solution:**

Let the three positive numbers be $x,y,z$ (where $x < y < z$)

Average of the three numbers = $\frac{x + y + z}{3}$

Acc. to ques, => $\frac{1}{3} \times (\frac{x + y + z}{3}) = z – 8$

=> $x + y + z = 9z – 72$

=> $x + y = 8z – 72$

Dividing both sides by 2, we get :

=> $\frac{x + y}{2} = 4z – 36$

Also, average of the lowest and the second lowest number is 8, => $\frac{x + y}{2} = 8$

=> $4z – 36 = 8$

=> $4z = 8 + 36 = 44$

=> $z = \frac{44}{4} = 11$

**Question 14: **If sum of smaller number x and two times the other number is equal to the sum of two times the smaller number and 16. The difference between the numbers is 6. Find the smaller number.

a) 4

b) 3

c) 6

d) 8

e) None of these

**14) Answer (A)**

**Solution:**

Let the smaller number be $x$ and the other number be $y$

Acc. to ques, => $x + 2y = 2x + 16$

=> $2y – x = 16$ ————-(i)

Also, difference between the numbers = $y – x = 6$ ————(ii)

Multiplying eqn(ii) by 2 and then subtracting it from (i), we get :

=> $(2y – 2y) + (-x + 2x) = 16 – 12$

=> $x = 4$

**Question 15: **The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 years and average age of males is 20 years. Find the number of males.

a) 8

b) 7

c) 10

d) 15

e) None of these

**15) Answer (D)**

**Solution:**

Let number of males = $x$

Average age of males = 20 years

=> Sum of age of males = $20 \times x = 20x$ years

Sum of age of females = 240 years

Acc. to ques, => $\frac{(20 x) + (240)}{15 + x} = 18$

=> $20x + 240 = 270 + 18x$

=> $20x – 18x = 2x = 270 – 240 = 30$

=> $x = \frac{30}{2} = 15$

**Question 16: **The difference between a two digit number and the number obtained by interchanging the two digits of the number is 9. If the sum of the two digits of the number is 15, then what is the original number?

a) 89

b) 67

c) 87

d) Cannot be determined

e) None of these

**16) Answer (C)**

**Solution:**

Let the unit’s digit of the number = $y$ and ten’s digit = $x$

=> Number = $10x+y$

Sum of digits = $x+y=15$ ———-(i)

According to ques, => $(10x+y)-(10y+x)=9$

=> $9x-9y=9$

=> $x-y=1$ ———-(ii)

Adding equations (i) and (ii), => $2x=15+1=16$

=> $x=\frac{16}{2}=8$

Substituting it in equation (i), => $y=15-8=7$

$\therefore$ Original number = 87

=> Ans – (C)

**Question 17: **Average score of a class of 50 students, in an exam is 34. Average score of the students who have passed is 52 and the average score of students who have failed is 16. How many students have failed in the exam?

a) 25

b) 20

c) 15

d) 18

e) 30

**17) Answer (A)**

**Solution:**

Let the number of students who failed = $x$

=> Number of students who passed = $(50 – x)$

Acc. to ques,

=> $\frac{(52 \times (50 – x)) + (16 \times x)}{50} = 34$

=> $52 \times 50 – 52x + 16x = 34 \times 50$

=> $52x – 16x = 50 \times (52 – 34)$

=> $36x = 50 \times 18$

=> $x = \frac{50 \times 18}{36} = 25$

**Question 18: **In a class, the average weight of 80 boys is 64 kg and that of 75 girls is 70 kg. After a few days, 60% of the girls and 30% of the boys leave. What would be the new average weight of the class (in kg)? Assume that the average weight of the boys and the girls remain constant throughout.

a) 63

b) 66.09

c) 68.5

d) 65.5

e) 57.5

**18) Answer (B)**

**Solution:**

Initially, number of boys = 80 and number of girls = 75

Average weight of boys = 64 kg and average weight of girls = 70 kg

Now, 60% of the girls and 30% of the boys leave

=> Boys left = $\frac{100 – 30}{100} \times 80 = 56$

Girls left = $\frac{100 – 60}{100} \times 75 = 30$

Since, average weight of the boys and the girls remains constant throughout

$\therefore$ New average weight of the class

= $\frac{(56 \times 64) + (30 \times 70)}{56 + 30} = \frac{3584 + 2100}{86}$

= $\frac{5684}{86} = 66.09$ kg

**Question 19: **The sum of a series of 5 consecutive odd numbers is 195. The second lowest number of this series is 9 less than the second highest number of another series of 5 consecutive even numbers. What is 40% of the second lowest number of the series of consecutive even numbers?

a) 16.8

b) 18.8

c) 19.4

d) 17.6

e) 16.4

**19) Answer (A)**

**Solution:**

Let the five consecutive odd numbers in increasing order = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$

Sum of these numbers = $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 195$

=> $5x = 195$

=> $x = \frac{195}{5} = 39$

Thus, the odd numbers are = 35 , 37 , 39 , 41 , 43

Let another series of even numbers in increasing order = $(y-4) , (y-2) , (y) , (y+2) , (y+4)$

Also, $37 = (y + 2) – 9$

=> $y = 37 + 9 – 2 = 44$

Thus, second lowest number of the even series = 44 – 2 = 42

$\therefore$ 40% of 42 = $\frac{40}{100} \times 42 = 16.8$

**Question 20: **P, Q and R have a certain amount of money with themselves. Q has 50% more than what P has, and R has ${1 \over 3}$rd of what Q has. If P,Q and R together have Rs. 246, then how much money does P alone have? (in Rs.)

a) 75

b) 60

c) 120

d) 82

e) 90

**20) Answer (D)**

**Solution:**

Let P has = $Rs. 100x$

=> Amount with Q = $100x + \frac{50}{100} \times 100x = Rs. 150x$

=> Amount with R = $\frac{1}{3} \times 150x = Rs. 50x$

Total amount together = $100x + 150x + 50x = 246$

=> $x = \frac{246}{300} = \frac{82}{100}$

=> $x = 0.82$

$\therefore$ Amount with P alone = $100 \times 0.82 = Rs. 82$