Free Algebra Subjective Test 01 Practice Test - 6th grade

Question 1

Express the following statement in the algebraic form "the sum of $3 x$ and $4 y$ is $8$ ." [1 MARK]

SOLUTION

Solution : Solution: 1 Mark

The algebraic form of the given statement is:
$3 x + 4 y = 8$

Question 2

Add: [2 MARKS]

(i) $3 x^{3} - 5 x^{2} + 8 x + 10$ ,
$15 x^{3} - 6 x - 23$ and
$9 x^{2} - 4 x + 15$

(ii) $3 a b^{2} - 2 b^{2} + a^{2}$ ,
$5 a^{2} b - 2 a b^{2} - 3 a^{2}$ and
$8 a^{2} - 5 b^{2}$

SOLUTION

Solution : Each Point: 1 Mark

(i) $(3 x^{3} - 5 x^{2} - + 8 x + 10) + (15 x^{3} - 6 x - 23) + (9 x^{2} - 4 x + 15)$

$= 3 x^{3} - 5 x^{2} + 8 x + 10 + 15 x^{3} - 6 x - 23 + 9 x^{2} - 4 x + 15$

Bringing like terms together, we get,

$= 3 x^{3} + 15 x^{3} - 5 x^{2} + 9 x^{2} + 8 x - 6 x - 4 x + 10 - 23 + 15$

$= 18 x^{3} + 4 x^{2} - 2 x + 2$

(ii) $(3 a b^{2} - 2 b^{2} + a^{2}) + (5 a^{2} b - 2 a b^{2} - 3 a^{2}) + (8 a^{2} - 5 b^{2})$

$= 3 a b^{2} - 2 b^{2} + a^{2} + 5 a^{2} b - 2 a b^{2} - 3 a^{2} + 8 a^{2} - 5 b^{2}$

Bringing like terms together, we get,

$= 3 a b^{2} - 2 a b^{2} - 2 b^{2} - 5 b^{2} + a^{2} - 3 a^{2} + 8 a^{2} + 5 a^{2} b$

$= 5 a^{2} b + a b^{2} + 6 a^{2} - 7 b^{2}$

Question 3

5x + 4y + 6pq - 2y + 3x + 4xy - 3pq - 2xy is equal to: [2 MARKS]

SOLUTION

Solution : Steps: 1 Mark
Answer: 1 Mark

Like terms can be added and subtracted.

So, $5 x + 3 x = 8 x$

Similarly, $4 y - 2 y = 2 y$

$4 x y - 2 x y = 2 x y$

$6 p q - 3 p q = 3 p q$
$\Rightarrow 5 x + 4 y + 6 p q - 2 y + 3 x + 4 x y - 3 p q - 2 x y$

= $8 x + 2 y + 2 x y + 3 p q$

Question 4

Rahul and his cousins spent ₹ (5x + 6y) for the clothes they bought, where ₹ x is the price of a pant and ₹ y is the price of a shirt. They wanted to return 2 shirts because the shirts were damaged and they wanted a refund. Then the money they spent on clothes will be equal to? [3 MARKS]

SOLUTION

Solution : Steps: 2 Marks
Answer: 1 Mark

Price of one shirt = ₹ y

So, price of two shirts = ₹ 2y

Rahul and his cousins will get ₹ 2y after giving back two shirts.

Now, the money they spent on clothes = (Money they spent before giving back the shirts) - (Cash they get back)

= (5x + 6y) - (2y)

Like terms can be subtracted.
$\Rightarrow 6 y$ and $2 y$ are like terms,

$∴$ The money they spent on clothes
= ₹ (5x + 6y - 2y)

= ₹ (5x + 4y)

Question 5

One-third of a number added to one-fifth gives 32. Find the number.[3 MARKS]

SOLUTION

Solution :
Steps: 2 Marks
Answer: 1 Mark

Let the number be $x$ .

One-third of the number $\Rightarrow \frac{1}{3} x$

One-fifth of the number $\Rightarrow \frac{1}{5} x$

Given: $\frac{1}{3} x + \frac{1}{5} x = 32$

$\Rightarrow \frac{5 x + 3 x}{15} = 32$

$\Rightarrow \frac{8 x}{15} = 32$

$\Rightarrow x = \frac{32 \times 15}{8} = 60$

$∴$ The required number = $60$

Question 6

A pattern is made using matchsticks. What is the number of matchsticks needed to create the next member in the pattern? [3 MARKS]

SOLUTION

Solution : Visualisation of question: 1 Mark
Identifying the pattern: 1 Mark
Solution: 1 Mark

Number of sticks used for the first member = 6
Number of sticks used for the second member = 6 + 5 = 11

Number of sticks used for the third member = 6 + 5 + 5 = 16
Therefore, number of sticks required for the fourth member = 16 + 5 = 21

Question 7

Find the value of: [4 MARKS]

(i) $p + q + 3 r$
$w h e n p = 1, q = 5, r = 2$

(ii) $2 a + 4 b + 5 c$
$w h e n a = 5, b = 10, c = 20$

(iii) $3 a - 2 b$
$w h e n a = 8, b = 10$

(iv) $5 x + 3 y - 6 z$ ,
$w h e n x = 3, y = 5, z = 4$

SOLUTION

Solution : Each option: 1 Mark

(i) $p + q + 3 r, w h e n p = 1, q = 5, r = 2$

$p + q + 3 r = 1 + 5 + 3 (2) = 12$

(ii) $2 a + 4 b + 5 c, w h e n a = 5, b = 10, c = 20$

$2 a + 4 b + 5 c = 2 (5) + 4 (10) + 5 (20)$

$= 10 + 40 + 100 = 150$

(iii) $3 a - 2 b, w h e n a = 8, b = 10$

$3 a - 2 b = 3 (8) - 2 (10)$

$= 24 - 20 = 4$

(iv) $5 x + 3 y - 6 z, w h e n x = 3, y = 5, z = 4$

$5 x + 3 y - 6 z = 5 (3) + 3 (5) - 6 (4) = 15 + 15 - 24$

$= 30 - 24 = 6$

Question 8

Translate each of the following statements into an equation: [4 MARKS]
a) The perimeter (p) of an equilateral triangle is three times of its side (a).
b) The diameter (d) of the circle is twice its radius (r).
c) The selling price (s) of an item is equal to the sum of the cost price (c) of an item and the profit (p) earned.
d) Amount (a) is equal to the sum of principal (p) and interest (i).

SOLUTION

Solution :
Each Option: 1 Mark

(a) Side of a equilateral triangle = a
According to question, p = 3a

(b) Diameter of circle = d
Radius of circle = r
According to question, d = 2r.

(c) Selling price = s, cost price = c and profit = p
According to question, s = c + p.

(d) Amount = a, Principal = p and Interest = i
According to question, a = p + i.

Question 9

(a) Sunita is half the age of her mother Geeta. Find their ages.
(i) After 4 years
(ii) Before 3 years

(b) A bus travels at 'v' km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using $v$ .

(c) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father's age?

[4 MARKS]

SOLUTION

Solution : (a) Solution: 2 Marks (1 Mark for Geeta and 1 mark for Sunita)
(b) Solution: 1 Mark
(c) Solution: 1 Mark

(a) Let the present age of Sunita’s mother Geeta = $2 x$ years

Present age of Sunita $= \frac{2 x}{2} = x y e a r s$

(i) After 4 years:
Age of Geeta $= (2 x + 4) y e a r s$
Age of Sunita $= (x + 4) y e a r s$

(ii) Before 3 years
Age of Geeta $= (2 x - 3) y e a r s$
Age of Sunita $= (x - 3) y e a r s$

(b) Speed of the bus = $v \frac{k m}{h r}$ i.e. (in 1 hr bus travels $v$ km )

$∴$ Distance travelled by bus after 5 hrs = $5 v k m$

Distance remaining to Beespur = $20 k m$

Distance from Daspur to Beespur
= Distance travelled by bus in 5 hrs + Distance remaining
= $(5 v + 20) k m$

(c) Let Sarita's present age be x years
$∴$ three times of Sarita's age =3x years
$∴$ Sarita's father age = 3x + 5 years

Question 10

Length and breadth of a bulletin board are $r c m$ and $t c m$ respectively. [4 MARKS]

(i) What will be the length (in cm) of the aluminium strip required to frame the board, if 10 cm extra strip is required to fix it properly.?

(ii) If $x$ nails are used to repair each board, how many nails will be required to repair 15 such board?

(iii) If $500 c m^{2}$ extra cloth per board is required to cover the edges, what will be the total area of the cloth required to cover 8 such boards?

(iv) What will be the expenditure for making 23 boards, if the carpenter charges Rs $x$ per board?

SOLUTION

Solution :
Each Point: 1 Mark

Perimeter of the bulletin board $= 2 (r + t) c m$
Area of the bulletin board $= r t c m^{2}$

(i) Length (in cm) of the aluminium strip required to frame the board = Perimeter + extra strip
= $2 (r + t) + 10 c m$

(ii) Repairing 1 board requires $x$ nails.
$∴$ Number of nails required to repair 15 boards = $15 x$

(iii) Total area of cloth required = Area of 8 bulletin boards + Area of extra cloth required for 8 boards
= ( $8 \times r \times t c m^{2}$ ) + ( $8 \times 500 c m^{2}$ )
= ( $8 r t + 4000) c m^{2}$

(iv) Carpenter's charge for making 1 board = Rs $x$
$∴$ Expenditure for making 23 boards = Rs $23 x$

Free Algebra Subjective Test 01 Practice Test - 6th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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