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Algebra Test - 11

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Algebra Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

What will be the expression after rationalising ?

A

B
2 +

C
4 +

D
4 –

E
None of the above

Solution

(By rationalising the denominator)
=
=
=
Question 2
1 / -0

Which of the following is the simplest form of the expression ?

A
8 +

B
8 +

C
8 +

D
10 +

E
2 +

Solution

(By rationalising the denominator)

=
=
= 8 +
Question 3
1 / -0

If 1 is taken as numerator and 13 + is taken as denominator, the expression will be equal to

A

B

C

D

E

Solution

(By rationalising the denominator)
Question 4
1 / -0

What will be the simplest form of the given expression?

A
x² + a² +

B
x + a +

C

D
x² + a² –

E
x – a +

Solution

(By rationalising the denominator)
Question 5
1 / -0

What will be the values of a and b, if ^.

A
a = 1, b = 2

B
a = 6, b = -2

C
a = 3, b = 3

D
a = 3, b = 2

E
a = 12, b = 15

Solution

(By rationalizing the denominator)

On comparing both sides, we get
a = 6, b = - 2
Question 6
1 / -0

The expression is equivalent to

A

B

C

D

E

Solution

= - 12 +
Question 7
1 / -0

The reciprocal of is

A

B

C

D

E

Solution

(By rationalising the denominator)
=
Question 8
1 / -0

Evaluate: (204)³

A
8189664

B
8489664

C
8489654

D
8289664

E
8479664

Solution

(204)³ = (200 + 4)³
= (200)³ + (4)³ + 3 (200) (4)² + 3 (200)² (4)
= 8000000 + 64 + 9600 + 480000
= 8489664
Question 9
1 / -0

If = ^{, find the value of n.}

A
3

B
4

C
5

D
6

E
7

Solution

= 7^{n + 3n - 1 - 2n - n - 1}
= 7^{n - 2}
Now,

3773 is divisible is by 11.
(3 + 7) - (3 + 7) = 0
= 343 = 7³
7^{n - 2} = 7³
n - 2 = 3
n = 5
Question 10
1 / -0

Solve the following equations for x and y:

2^{x - 1} + 3^{y - 1} = 5 and 2^{x + 3} - 3^y = 23

A
x = 2, y = 2

B
x = 3, y = 2

C
x = 2, y = 7

D
y = 3, x = 1

E
x = 2, y = 5

Solution

2^{x - 1} + 3^{y - 1} = 5 3^{y - 1} = 5 - 2^{x - 1}
2^{x + 3} - 3^y = 23 2^{x + 3} - (3^{y - 1}) 3 = 23
2^{x + 3} - 3 (5 - 2^{x - 1}) = 23
2^{x + 3} - 15 + 3 . 2^{x - 1} = 23
2^x = 23 + 15
2^x = = 2²
x = 2
3^{y - 1} = 5 - 2^{2 - 1} = 3
y - 1 = 1
y = 2
Question 11
1 / -0

Find the value of , if log_a961 = 1.

A
1

B
2

C
3

D
4

E
5

Solution

=

= =

log_a961 = log_a961
= = = log_a31
Now, log_a31 = 1
a = 31
So, (a + 1) = 32
(a + 1)^(1/5)= 32⁽^1/5) = 2
Question 12
1 / -0

Evaluate:

A

B

C

D

E

Solution

=

=

= ₌
=

=
Therefore, =
Question 13
1 / -0

Area of a right angled isosceles triangle is increased by 21% of its actual area. If it is bordered by 3 metre wide strip, find the length of equal sides.

A
35 m

B
32 m

C
30 m

D
28 m

E
None of these

Solution

Let x be length of each of the two equal sides.
Then, area = x²When triangle is bordered, area = (x + 3)²
According to the statement, (x + 3)² = x² + 21% of
(x + 3)² =
x + 3 =
x = 30 m
Question 14
1 / -0

If x + = and y + = , then find the value of x³y³ + .

A
0

B
-118

C
-126

D
121

E
138

Solution

x + = and y + =
Multiplying,
xy + x × + y ×+ × = ×xy + = 2
Now, x³y³ + = (xy)³ +
= – 3xy ×
= 2³ – 3×21(2)
= 8 – 126
= –118
Question 15
1 / -0

What is the difference of from the sum of , and ?

A
1/5x

B
2x

C
1

D
0

E
None of these

Solution

+ –
= –
= –
=
=
=
Now, – = 0
Question 16
1 / -0

Solve: + 4 = and 9 - = +

A
,

B
,

C
,

D
,

E
,

Solution

4x - 6y + 12 = 3 []
8x - 12y + 24 = 3x + 2y
5x - 14y = - 24………(1)
And
=
270 - 12x + 18y = 20x + 15y
32x - 3y = 270 ……..(2)

(1) ´ 3 15x - 42y = - 72
(2) ´ 14 448x - 42y = 3780
On subtracting,
433x = 3852
x =
y =
=
=
=
Question 17
1 / -0

Evaluate: x-2 - y-2, where x = 4 + and y = 12 - 3

A
7

B

C
7

D
8

E
8

Solution

x2 = 16 + 8 + 3
= 19 + 8
y² = (12 - 3)²
= 144 + 27 - 72
= 171 - 72
= 9(19 - 8)
x^-2 - y^-2 = -

=
=
=
=
Question 18
1 / -0

Ram said to Sham, If you give me half of your money, I will have four times of yours, and if I give you of what I have, you will have 130 more than me. Find the total money with Ram and Sham.

A
Rs. 250

B
Rs. 350

C
Rs. 400

D
Rs. 300

E
Rs. 200

Solution

Let Ram had x rupees and Sham had y rupees.
According to the first case:
x + = 4........(1)
According to the second case:
x + 130 = y + x........(2)
From (1) and (2),
x = 150
y = 100
Question 19
1 / -0

What is the value of x² + when x - = 4?

A
18

B
36

C
16

D
30

E
20

Solution

x - = 4
By squaring both sides, we get
x² + - 2 = 16
So, x² + = 18
Question 20
1 / -0

If (1/24)^x = (1/3)^a(1/2)^b, find the relation between 'a' and 'b'.

A
3a = b

B
a = 3b

C
3a = 4b

D
3a = 2b

E
2a = 3b

Solution

(1/24)^x= (1/3)^a(1/2)^b
Given,
24^x = 3^a 2^b.........(Taking reciprocal of both the sides)
2^3x. 3^x = 3^a 2^b
a = x and b = 3x
Question 21
1 / -0

Find the value of (128)³.

A
2587142

B
2587162

C
2097162

D
2085152

E
2097152

Solution

(128)³ = (100 + 28)³
= 100³ + 28³ + 3.100.28² + 3.100².28
= 1000000 + 21952 + 235200 + 840000
= 2097152
Question 22
1 / -0

If 4x - 6y = 7 and = , what is the value of (5x + 3y)?

A

B

C

D

E

Solution

4x - 6y = 7 ........(1)
=
7y = 6x - 6y
6x - 13y = 0 .......(2)
Now, (1)6 and (2)4 and subtracting (2) from (1), we get
24x - 36y - 24x + 52y = 42 - 0
16y = 42
y =
Substituting y = 42/16 in (1), we get

4x = 7 + 6y
4x = 7 + 6
=
=
x =
5x + 3y = [5] +[3]
=
=
Question 23
1 / -0

If 6 + = , what is the value of x?

A

B

C

D

E
None of these

Solution

6 + = = 6 +
=
1 + = = 1 +
=
1 + = = 1 +
x = 4/3
Question 24
1 / -0

If = , what is the value of ?

A
-

B

C
-

D

E
-

Solution

=
18a + 14b = 12a + 15b
6a = b
Therefore,
=
=
=

= -
Question 25
1 / -0

A shop owner had a certain amount of sugar. He sold of the sugar plus 1 kg to the 1^st party, of the remaining sugar to the 2^nd party and of the remaining sugar plus 1 kg to the third party. At last, he was left with 130 kg of sugar. What was the quantity of sugar that the shop owner had initially?

A
384 kg

B
438 kg

C
273 kg

D
504 kg

E
612 kg

Solution

Let the total amount of sugar be x kg.
Then, x - = (x - 1) kg of sugar is left after the 1^stsale.
So, - = (x - ) kg of sugar is left after the 2^nd sale.
Finally, (x - ) - 1 kg of sugar is left after the final sale.
So, x - - x + - 1 = 130
= 130 + + 1 -
= 131 + -
=
= x = ×
x = 438 kg

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