Algebra Test

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Algebra Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The method of finding solution by trying out various values for the variable is called

A
Error method

B
Trial and error method

C
Testing method

D
Checking method

Solution

The method of finding solution by trying out various values for the variable is Trial and error method
Question 2
1 / -0

Meera bought packs of trading cards that contain 10 cards each. She gave 7 cards to her friend.

x = Number of packs of trading cards

Which expression shows the number of cards left with Meera?

A
10x − 7

B
7x − 8

C
5 − 10x

D
8 − 5x

Solution

Number of cards in pack = 10 ∴ Number of cards in x packs = 10x ∴ Number of cards left with Meera = 10x − 7
Question 3
1 / -0

Which of the following equations does not have a solution in integers?

A
x + 1 = 1

B
x − 1 = 3

C
2x + 1 = 6

D
1 − x = 5

Solution

(a) x + 1 = 1

⇒x+1−1=1−1

[Subtracting 1 from both sides]

⇒x=0, which is an integer

(a) x−1=3

⇒x−1+1=3+1

[Adding 1 to both sides]

⇒x=4, which is an integer.

(c) 2x+1=6

⇒2x+1−1=6−1

[Subtracting 1 from both sides]

⇒2x=5

⇒2x2=52. [Dividing both sides by 2]

⇒x=52, which is not an integer.

(d) 1−x=5

⇒1−x−1=5−1

[Subtracting 1 from both sides]

⇒−x=4

⇒−(−x)=−4

[Multiplying both sides by (-1)]

⇒x=−4, which is an integer.

Thus, the above conditions show that equation 2x+1=6 does not have a solution in integers.
Question 4
1 / -0

When Raju multiplies a certain number by 17 and adds 4 to the product, he gets 225. Find the number.

A
13

B
14

C
15

D
15

Solution

Let the numbers be x. According to question, 17x + 4 = 225
⇒ 17 x = 225 − 4 = 221 ⇒ x = 221/17 = 13
Question 5
1 / -0

Preeti travelled 3x km distance by walk, 9y km by cycle and 5 km by bus. The total distance covered by Preeti is __________.

A
(3x − 9y + 5) km

B
(3x + 9y + 5) km

C
(3x − 9y − 5) km

D
(9x + 3y − 5) km

Solution

Total distance covered by Preeti = (3x + 9y + 5) km
Question 6
1 / -0

The algebraic expression for the statement "Product of x and reciprocal of a, subtracted from the product of y and reciprocal of b" is

A
y^b − x^a

B
y − x/a − b

C
xa − yb

D
1/yb − xa

Solution

The algebraic expression for the statement "Product of x and reciprocal of a, subtracted from the product of y and reciprocal of b" is y^b − x^a
Question 7
1 / -0

I had Rs.350 with me. I gave Rs. (x/2) to Amit, Rs.(x/3) to Shreya and I am left with Rs.(x/3). The amount I gave to Amit is ________.

A
Rs.150

B
Rs.100

C
Rs.250

D
Rs.200

Solution

According to question, we have x/2 + x/3 + x/3 = Rs.350 ⇒ 3x+2x+2x/6 = Rs.350 ⇒ 7x=6×350
=Rs.2100 ⇒ x = 21007
=Rs.300 Now, amount given to Amit =Rs.(x/2) = Rs.(300/2) = Rs.150
Question 8
1 / -0

If five times of a number is 80, then the number is ________.

A
-16

B
1

C
16

D
40

Solution

Let the number be x. According to question, 5x = 80 ⇒ x = 16
Question 9
1 / -0

The equation for the statement; 'Half of a number added to 10 is 15' is _______.

A
x/2 = 10 + 5

B
x/2 + 10 = 15

C
x/2 + 15 = 10

D
x/2 = 10 + 15

Solution

Let the number be x. ∴x/2 + 10 = 15

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Algebra Test - 2

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Algebra Test - 2

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Question 1

Error method

Trial and error method

Testing method

Checking method

Solution

Question 2

10x − 7

7x − 8

5 − 10x

8 − 5x

Solution

Question 3

x + 1 = 1

x − 1 = 3

2x + 1 = 6

1 − x = 5

Solution

Question 4

13

14

15

15

Solution

Question 5

(3x − 9y + 5) km

(3x + 9y + 5) km

(3x − 9y − 5) km

(9x + 3y − 5) km

Solution

Question 6

yb − xa

y − x/a − b

xa − yb

1/yb − xa

Solution

Question 7

Rs.150

Rs.100

Rs.250

Rs.200

Solution

Question 8

-16

1

16

40

Solution

Question 9

x/2 = 10 + 5

x/2 + 10 = 15

x/2 + 15 = 10

x/2 = 10 + 15

Solution

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y^b − x^a