6th Grade > Mathematics
ALGEBRA MCQs
Total Questions : 96
| Page 9 of 10 pages
Answer: Option D. ->
34 units
:
D
When y=5
Distance=(2y2+y+2)=2×52+5+2=50+5+2=57 units
When y =3,
Distance =(2y2+y+2)=2×32+3+2=18+3+2=23 units
∴ Required difference = 57 – 23 = 34 units.
:
D
When y=5
Distance=(2y2+y+2)=2×52+5+2=50+5+2=57 units
When y =3,
Distance =(2y2+y+2)=2×32+3+2=18+3+2=23 units
∴ Required difference = 57 – 23 = 34 units.
Answer: Option C. ->
12
:
C
:
C
x−3=9Add 3 on both sides,⇒x−3+3=9+3⇒x+0=12⇒x=12
Answer: Option B. ->
4xy,3yx
:
B
Like terms are terms that contain the same variables with the same power.
Here, 3yx and 4xy are like terms,
3yx and 8x2 are unlike terms,
4xy and 8x2 are unlike terms.
:
B
Like terms are terms that contain the same variables with the same power.
Here, 3yx and 4xy are like terms,
3yx and 8x2 are unlike terms,
4xy and 8x2 are unlike terms.
Answer: Option B. ->
₹ (13x - 1)
:
B
Total distance travelled = x km
So, the distance applicable for additional charge is (x−2) km.
∴ Total additional charge =₹13(x−2)
Total charge = Fixed charge + Additional charge
=13(x−2)+25
=13x−26+25
=₹(13x -1)
∴ Expression for total charge is ₹(13x -1).
:
B
Total distance travelled = x km
So, the distance applicable for additional charge is (x−2) km.
∴ Total additional charge =₹13(x−2)
Total charge = Fixed charge + Additional charge
=13(x−2)+25
=13x−26+25
=₹(13x -1)
∴ Expression for total charge is ₹(13x -1).
Answer: Option A. ->
2n+1
:
A
Even numbers are divisible by 2.
So, all even numbers can be represented by 2 × n; where n represents whole numbers 0, 1, 2....
All, even number are succeeded by an odd number.
∴ 2n+1 gives the pattern formed by odd numbers.
For n=0, 2n+1 = 2 × 0 + 1 = 1
For n=1, 2n+1 = 2 × 1 + 1 = 3
For n=2, 2n+1 = 2 × 2 + 5 = 5.
So, it is clear that A is the correct option.
:
A
Even numbers are divisible by 2.
So, all even numbers can be represented by 2 × n; where n represents whole numbers 0, 1, 2....
All, even number are succeeded by an odd number.
∴ 2n+1 gives the pattern formed by odd numbers.
For n=0, 2n+1 = 2 × 0 + 1 = 1
For n=1, 2n+1 = 2 × 1 + 1 = 3
For n=2, 2n+1 = 2 × 2 + 5 = 5.
So, it is clear that A is the correct option.
Answer: Option C. ->
10x+6y
:
C
:
C
Adding the coefficients of x and y, we get
3x+4y +
7x+2y
_______
10x+6y
3x+4y+2y+7x=10x+6y
Answer: Option C. ->
10x+6y
:
Solution: 1 Mark
The algebraic form of the given statement is 2x+28=45
:
Solution: 1 Mark
The algebraic form of the given statement is 2x+28=45
Answer: Option C. ->
10x+6y
:
Solution: 1 Mark each
(i) (4a−b+6c)−(3a−4b+5c)
= 4a−b+6c−3a+4b−5c
= 4a−3a−b+4b+6c−5c
= a+3b+c
(ii) (a−4b−2c)−(5a−3b+2c)
= a−4b−2c−5a+3b−2c
= a−5a−4b+3b−2c−2c
= −4a−b−4c
:
Solution: 1 Mark each
(i) (4a−b+6c)−(3a−4b+5c)
= 4a−b+6c−3a+4b−5c
= 4a−3a−b+4b+6c−5c
= a+3b+c
(ii) (a−4b−2c)−(5a−3b+2c)
= a−4b−2c−5a+3b−2c
= a−5a−4b+3b−2c−2c
= −4a−b−4c
Answer: Option C. ->
10x+6y
:
Equation: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Let the number be x.
The number decreased by 15 ⇒(x−15)
The number so obtained is multip[lied by 3 ⇒3×(x−15)
Given: 3(x−15)=81⇒3x−45=81
⇒3x=81+45
⇒x=1263=42
∴ The required number = 42
:
Equation: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Let the number be x.
The number decreased by 15 ⇒(x−15)
The number so obtained is multip[lied by 3 ⇒3×(x−15)
Given: 3(x−15)=81⇒3x−45=81
⇒3x=81+45
⇒x=1263=42
∴ The required number = 42
Answer: Option C. ->
10x+6y
:
:
Concept: 1 Mark
Solution: 1 Mark
Total score of Kimmy =2a+5b+6d
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