**Question 1. If 4x +**

^{3}/_{5}= 5, then x =Answer: A)

=> 4x = 5-

=> 4x =

=> x =

x =

^{11}/_{10}**Solution:-**4x +^{3}/_{5}= 5=> 4x = 5-

^{3}/_{5}=> 4x =

^{22}/_{5}=> x =

^{22}/_{5}÷ 4x =

^{11}/_{10}**Question 2. If**

^{x}/_{3}-^{5}/_{2}= 6, then x = ?^{51}/

_{2}

^{52}/

_{3}

^{53}/

_{4}

^{54}/

_{5}

Answer :

=>

=>

=> x =

=> x =

^{51}/_{2}**Solution:-**^{x}/_{3}-^{5}/_{2}= 6=>

^{x}/_{3}= 6 +^{5}/_{2}=>

^{x}/_{3}=^{17}/_{2}=> x =

^{(17 * 3)}/_{2}=> x =

^{51}/_{2}**Question 3. The age of father is 3 times the age of the age of son. If sum of their ages is 48 yeras, then the age of father ans son are (in years):**

Answer : A) Father=36, Son=12

Given, x + 3x = 48

=> 4x = 48

=> x = 12

Son age = 12

∴ Fathers age = 36

**Solution:-**let the age of son be x,Given, x + 3x = 48

=> 4x = 48

=> x = 12

Son age = 12

∴ Fathers age = 36

**Question 4.Length of a rectangle is 8m less than twice of breadth. If the perimeter of the rectangle is 56m. Find its length and breadth**

Answer: A) Length = 16m and breadth = 12m

=>2(2m-8)+4m=56

=>4m-8+4m=56

=>8m-8=56

=>8m=48

=>m=6

**Solution:-**2{(2m-8)+2m}=56=>2(2m-8)+4m=56

=>4m-8+4m=56

=>8m-8=56

=>8m=48

=>m=6

**Question 5. The sum of a two digit and the number obtained by interchanging the digits of the number is 121.If the digis differ by 5, Then find the number.**

Answer: A) 38,83

∴ Number = 10x + y

On interchanging the digits 10y + x

=>10x + y + 10y + x = 121 (Equation 1)

Since, digits differ by 5

∴ x = (y +5)

Now putting value of x in equation 1

=>10(y+5) + y + 10y + y + 5 = 121

=>10y + 50 + y + 10y + (y + 5) = 121

=>22y + 55 = 121

=>22y = 66

=>y = 3

∴ x = 3 + 5 = 8

∴ Number is = 83

**Solution:-**Let a number have x at ten's place and y at one's place.∴ Number = 10x + y

On interchanging the digits 10y + x

=>10x + y + 10y + x = 121 (Equation 1)

Since, digits differ by 5

∴ x = (y +5)

Now putting value of x in equation 1

=>10(y+5) + y + 10y + y + 5 = 121

=>10y + 50 + y + 10y + (y + 5) = 121

=>22y + 55 = 121

=>22y = 66

=>y = 3

∴ x = 3 + 5 = 8

∴ Number is = 83

**Question 6. The age of Mira, Tina and Saina are in the ratio 6:4:7 respectively, if the sum of their ages is 34 years, what is Sania's age?**

Answer: D) 14 years

Given, 6x + 4x + 7x = 34

=>17x = 34

=>x = 2

∴ Mira's age = 6*2 = 12 years

∴ Tina's age = 4*2 = 8 years

∴ Sania's age = 7*2 = 14 years

**Solution:-**Let the age of Mira, Tina and Saina are in the ratio 6x, 4x, 7x respectively,Given, 6x + 4x + 7x = 34

=>17x = 34

=>x = 2

∴ Mira's age = 6*2 = 12 years

∴ Tina's age = 4*2 = 8 years

∴ Sania's age = 7*2 = 14 years

**Question 7. (**

^{3}/_{4})^{th}of a number is 20 more than half of the same number. The required number is ______.Answer: D) 80

Given, (

=>

=>

=> x/4 = 20

=> x = 20*4

=> x = 80

**Solution:-**Let the number be x.Given, (

^{3}/_{4})x = (^{x}/_{2}) + 20=>

^{3x}/_{4}-^{x}/_{2}= 20=>

^{3x - 2x}/_{4}= 20=> x/4 = 20

=> x = 20*4

=> x = 80

**Question 8. In a class of 100 students, 55 students have passed in Mathematics and and 67 students have passed in Physics. Then, the number of students who have passed in Physics only is.**

Answer: D) 45

Total number of students = 100

Total number of students passed in mathematics = 55

No of students who only passed by Physics is = x

=> x = 100-55

=> x = 45

**Solution:-**Let the number of students passed by Physics be x,Total number of students = 100

Total number of students passed in mathematics = 55

No of students who only passed by Physics is = x

=> x = 100-55

=> x = 45

**Question 9. If, 8x + 3 = 27 + 2x. Then x =?**

Answer: B) 4

=> 6x + 3 = 27

=> 6x = 24

=> x = 4

**Solution:-**8x + 3 = 27 + 2x=> 6x + 3 = 27

=> 6x = 24

=> x = 4

**Question 10. If 2x - 3 = x + 2, then x =?**

Answer: C) 5

=> (2x-x) - 3 = 2

=> x - 3 = 2

=> x = 5

**Solution:-**2x - 3 = x +2=> (2x-x) - 3 = 2

=> x - 3 = 2

=> x = 5