## RS Aggarwal Class 7 solutions of Chapter 7 Linear Equations in One Variable CCE Test Paper in pdf free download

**Question 1.****Evaluate x**^{3} + y^{3} + z^{3}** – 3xyz when x = -2, y = -1 and z = 3.****Solution:**

We have:

x^{3} + y^{3} + z^{3} – 3xyz

= (-2)^{3} + (-1)^{3} + (3)^{3} – 3 x (-2) x (-1) x 3

= – 8 – 1 + 27 – 18 = -27 + 27 = 0

**Question 2.****Write the coefficient of* in each of****the following:****(i) -5xy****(ii) 2xy**^{2 }**z****(iii) abx****Solution:**

Co-efficient of x in the given numbers are

(i) -5y

(ii) 2y^{2 }z

(iii) ab

**Question 3.****Subtract x**^{2} – 2xy + 5y^{2} – 4 from 4xy – 5x^{2} – y^{2}** + 6.****Solution:**

We have:

(4xy – 5x^{2} – y^{2} + 6) – (x^{2} – 2xy + 5y^{2} – 4)

= 4xy – 5x^{2} – y^{2} + 6 – x^{2} + 2xy – 5y^{2} + 4

= -6x^{2} – 6y^{2} + 6xy +10

= -2 (3x^{2} + 3y – 3xy – 5)

**Question 4.****How much less is x**^{2} – 2xy + 3y^{2} than 2x^{2} – 3y^{2}** + xy ?****Solution:**

We have:

(2x^{2} – 3y^{2} + xy) – (x^{2} – 2xy + 3y^{2})

= 2x^{2} – 3y^{2} +xy – x^{2} + 2xy – 3y^{2}

= 2x^{2} – x^{2} – 3y^{2} – 3y^{2} + xy + 2xy

= x^{2} – 6y^{2} + 3xy

x^{2} – 2xy + 3y^{2} is less than 2x^{2} – 3y^{2} + xy by x^{2} – 6y + 3xy.

**Question 5.****Find the product**

**Solution:**

We have:

**Question 6.****Simplify:****(3a + 4) (2a – 3) + (5a – 4) (a + 2).****Solution:**

We have:

(3a + 4) (2a – 3) + (5a – 4) (a + 2)

= {3a (2a – 3) + 4 (2a – 3)} + {5a (a + 2) – 4 (a + 2)}

= (6a^{2} – 9a + 8a – 12) + (5a^{2} + 10a – 4a – 8)

= (6a^{2} – a – 12) + (5a^{2} + 6a – 8)

= (11a^{2} + 5a – 20)

**Question 7.****Solution:**

**Question 8.****Solve : 0.5x + = 0.25x + 7.****Solution:**

We have:

**Question 9.****The sum of two consecutive odd numbers is 68. Find the numbers.****Solution:**

Let the consecutive odd number be

x and (x + 2)

x + (x + 2) = 68

2x + 2 = 68

2x = 68 – 2 = 66

x = 33

The required numbers are 33 and (33 + 2), i.e., 35.

**Question 10.****Reenu’s father is thrice as old as Reenu. After 12 years he will be just twice his daughter. Find their present ages.****Solution:**

Let Reenu’s present age be x

Then, her father’s present age will be 3x

Reenu’s age after 12 years = (x + 12)

3x + 12 = 2x + 24

x = 12

Reenu’s present age = x = 12 yrs

And her father’s age = 3x = (3 x 12) = 36 yrs

**Mark (✓****) against the correct answer in each of the following :****Question 11.****Solution:**

**Question 12.****Solution:**

**Question 13.****Solution:**

**Question 14.****A number when multiplied by 4 is increased by 54. The number is****(a) 21****(b) 16****(c) 18****(d) 19****Solution:**

(c) 18

Let the number be x.

According to the equation, we have:

4x = x + 54

⇒ 3x = 54

⇒ x = 18

**Question 15.****Two complementary angles differ by 14°. The larger angle is****(a) 50°****(b) 52°****(c) 54°****(d) 56°****Solution:**

(b) 52°

Let the two complementary angles be x° and (90 – x)°.

According to the equation, we have:

x – (90 – x) = 14

⇒ 2x = 104

⇒ x = 52

(90° – x)° = 90° – 52° = 38°

The larger angle is 52°.

**Question 16.****The length of a rectangle is twice its breadth and its perimeter is 96 m. The length of the rectangle is****(a) 28 m****(b) 30 m****(c) 32 m****(d) 36 m****Solution:**

(c) 32 m

Let the length and breadth of the rectangle be l m and b m, respectively.

According to the questions, we have:

l = 2b ……(i)

2 (l + b) = 96 …..(ii)

Now, 2 (2b+ b) = 96

⇒ 6b = 96

⇒ b = 16

Length = 16 x 2 m = 32 m

**Question 17.****The ages of A and B are in the ratio 4 : 3. After 6 years their ages will be in the ratio 11 : 9. A’s present age is****(a) 12 years****(b) 16 years****(c) 20 years****(d) 24 years****Solution:**

(b) 12 years

Let the ages of A and B be x and y years respectively,

**Question 18.****Fill in the blanks.****(i) -2a**^{2}** b is a ………..****(ii) (a**^{2} – 2b^{2}**) is a …………****(iii) (a + 2b – 3c) is a …………..****(iv) In -5ab, the coefficient of a is ………….****(v) In x**^{2}** + 2x – 5, the …………….. term is -5.****Solution:**

(i) -2a^{2} b is a monomial.

(ii) (a^{2} – 2b^{2}) is a binomial.

(iii) (a + 2b – 3c) is a trinomial.

(iv) In -5ab, the coefficient of a is -5.

(v) In x^{2} + 2x – 5, the constant term is -5.

**Question 19.****Write ‘T for true and ‘F’ for false for each of the following :****(i) In -x, the constant term is -1.****(ii) The coefficient of x in x**^{2}** – 3x + 5 is 3.****(iii) (5x – 7) – (3x – 5) = 2x – 12.****(iv) (3x + 5y) (3x – 5y) = (9x**^{2} – 25y^{2}**).****(v) If a = 2 and b = , then the value of ab (a**^{2} + b^{2}**) is 4.****Solution:**

(i) False.

The coefficient of x is -1.

(ii) False.

The coefficient of x in – 3.

(iii) False.

LHS = (5x – 7) – (3x – 5) = 5x – 7 – 3x + 5 = 2x – 2.

(iv) True.

LHS = (3x + 5y) (3x – 5y)

= 3x (3x – 5y) + 5y (3x – 5y)

= 9x^{2} – 15xy + 15xy – 25y^{2}

= 9x^{2} – 25y^{2}

(v) True

(a^{2} + b^{2})

## Leave a Reply