Algebra(6th Grade > Mathematics ) Questions and answers for exam Preparation

6th Grade > Mathematics

ALGEBRA MCQs

Total Questions : 96 | Page 5 of 10 pages

Question 41. One-third of a number added to one-fifth gives 32. Find the number.[3 MARKS]

Discuss Question

:
Steps: 2 Marks
Answer: 1 Mark
Let the number be

x

.
One-thirdof 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 42. 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?

Discuss Question

:
Each Point: 1 Mark
Perimeter of the bulletinboard

= 2 (r + t) c m

Area of thebulletinboard

= 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

Question 43. 5x + 4y + 6pq - 2y + 3x + 4xy - 3pq - 2xy is equal to: [2 MARKS]

Discuss Question

:
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 44.
A pattern is made using matchsticks. What is the number of matchsticks needed to create the next member in the pattern? [3 MARKS]

Discuss Question

:
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 45. (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]

Discuss Question

:
(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

∴

three times of Sarita's age =3x years

∴

Sarita's father age = 3x + 5 years

Question 46. 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}

Discuss Question

:
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 47.

Value of $2 x y$ when $x = 4$ and $y = 0$ is _________.

Discuss Question

Answer: Option A. -> 0
:
A

$2 x y = 2 \times x \times y$
On substituting the values of $x$ and $y$ , we get

$2 x y = 2 \times 4 \times 0 = 8 \times 0 = 0$

Question 48.

A number when multiplied by 2 and decreased by 41 gives 3. Find the number.

Discuss Question

Answer: Option C. -> 22
:
C

Let the number be $x$ .
$∴ 2 x - 41 = 3 \Rightarrow 2 x = 44 \Rightarrow x = \frac{44}{2} = 22$

Question 49.

The coefficient of term $x y^{2}$ in expression $4 x y^{2} + x^{3} - x^{5} + 2 x y^{2}$ is ______.

Discuss Question

Answer: Option D. -> 6
:
D
Since

4 x y^{2}

and

2 x y^{2}

are like terms, the given expression can be written as follows.

4 x y^{2} + x^{3} - x^{5} + 2 x y^{2} = 6 x y^{2} + x^{3} - x^{5}

Therefore, the coefficient of

x y^{2}

is 6.

Question 50.

Identify the binomial(s) among the given expressions.

7x+8xy
1+2x+3y
3x+7xy+yx
3x+2y

Discuss Question

Answer: Option D. -> 3x+2y
:
A, C, and D
7x +8 xy has two unlike terms. So, it is a binomial.
1 + 2x +3y has three unlike terms. So, it is a trinomial.
3x + 7xy + yx = 3x + 8xy which has two unlike terms. So, it is a binomial.
3x + 2y has two unlike terms. So, it is a binomial.