Mix Test

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Mix Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

If x = – 19, then which of the following numbers is the value of y such that the equation 9x – 5y = 19 is true?

A
– 36

B
– 38

C
38

D

Solution

9x – 5y = 19
x = – 19
9 (– 19) – 5y = 19
– 5y = 19 – 9(– 19)
– 5y = 19(1 + 9)
5y = – 19 10
y = – 38
Question 2
1 / -0

Solve |x² + x| – 5 < 0.

A
x < 0

B
x > 0

C
x belongs to the set of all real numbers

D
None of these

Solution

|x² + x| – 5 < 0
Put x = 1,
1 + 1 – 5 < 0
–3 < 0, which is true.
Put x = 3,
(3² + 3) – 5 < 0
12 – 5 < 0
7 < 0, which is not true.
x = 1 satisfies the relation with x, but x = 3 does not satisfy the given relation.
Putting x = –1, we get (1 – 1) - 5 < 0, which is true.
Putting x = –4, we get (16 – 4) – 5 < 0, which is not true.
Options (1), (2) and (3) can be eliminated.
Hence, option (4) is the correct answer.
Question 3
1 / -0

The number of real roots of the equation (x - 1)² + (x - 2)² + (x - 3)² = 0 is

A
1

B
2

C
3

D
None of these

Solution

(x - 1)² + (x - 2)² + (x - 3)² = 0
x² + 1 - 2x + x² + 4 - 4x + x² + 9 - 6x = 0
3x²- 12x + 14 = 0
For given equation to have real roots, D ≥ 0.
Here, D = (-12)² - 4 (14) (3)
= 144 - 168
= -24 < 0
No real roots.
Question 4
1 / -0

The slope of a line is not defined if the line is

A
parallel to x-axis

B
parallel to the line x – y = 0

C
perpendicular to y-axis

D
parallel to y-axis

Solution

The answer is option (4).
Question 5
1 / -0

The area of a square is 196 cm². What will be the radius of the largest semicircle that can be drawn completely inside on one of its sides?

A
98 cm

B
14 cm

C
7 cm

D
28 cm

Solution

If the area of a square is 'a²', then the length of one of its sides is 'a'. The radius (r) of a semicircle drawn completely inside on one of its sides is equal to 'a/2'.

Thus, the area of the given square = a² = 196 cm²
a =
So, the length of one side of the square = 14 cm
And the radius of the given semicircle = a/2 = 14/2 = 7 cm
Question 6
1 / -0

The solution set of 3(2x + 4) > 4x + 10 is

A
x = - 1

B
x > - 1

C
x < 1

D
x - 1

Solution

3(2x + 4) > 4x + 10
6x + 12 > 4x + 10
2x + 12 > 10
2x > - 2
x > - 1
Question 7
1 / -0

If two lines are at 90^o, then the product of the slope is

A
- 1

B
0

C
1

D
Not fixed

Solution

Let m₁ be the slope of 1st line and m₂ be the slope of 2nd line
We know that, if θ be the angle between these two lines, then

Given that θ = 90^o
So,

Hence, if two lines are at 90^o, then the product of the slope of line is - 1.
Question 8
1 / -0

If A, B and C denote the areas of three adjacent faces of a rectangular solid and V denotes its volume, then

A
V² = 2ABC

B
V² = ABC

C
V² = 8ABC

D
V² = 9ABC²

Solution

Area of one face = l × h = A
Area of second face = h × b = B
Area of third face = l × b = C
By multiplying them, we get
= l² b² h² = ABC … (1)
Volume of rectangular solid = l × b × h
V = l b h
Taking square both side, V²= l² b²h² ...(2)
By comparing the equations (1) and (2), we get
V² = ABC
Coreect option is 2
Question 9
1 / -0

A person earns Rs. 2048.45 in January and Rs. 2892.48 in February. He spends Rs. 4800 out of that money at the end of two months. What is the balance with him?

A
Rs. 183.83

B
Rs. 163.93

C
Rs. 159.13

D
Rs. 140.93

Solution

Money earned in January = Rs. 2048.45
Money earned in February = Rs. 2892.48
Total money earned = Rs. 2048.45 + Rs. 2892.48 = Rs. 4940.93
He spends Rs. 4800 at the end of two months, so balance = Rs. 4940.93 - Rs. 4800 = Rs. 140.93
Question 10
1 / -0

Ametre wide rectangular plank is placed symmetrically on the diagonal of a square of side 8 metres as shown in the figure.

The area of the plank is

A
(16- 3) sq.m

B
7 sq.m

C
98 sq.m

D
14 sq.m

Solution

AL = AP = x (say), PL =

x² + x² = 2
2x² = 2
∴ x = 1
LB = AB - AL = 8 - 1 = 7
Similarly, MB = 7
∴LM = = 7The area of the plank = 7 × = 14 m²
Question 11
1 / -0

In the given figure, || m and n is a transversal. If 7 = 100°, then the measure of 2 is

A
100°

B
10°

C
80°

D
160°

Solution

Given: 7 = 100°

If a transversal intersects two parallel lines, then the sum of interior angles on the same side of the transversal is 180°.
|| m and n is a transversal.
So, 7 + 2 = 180°(∵Sum of interior angles on the same side of a transversal is 180°)
100° + 2 = 180°
2 = 80°
Question 12
1 / -0

Due to inflation, the price of an item is increased by 25%, but at the end of the year, it decreased by 20% and became the same. What was the original price?

A
Rs. 200

B
Rs. 350

C
Rs. 360

D
Any value of original price

Solution

Let x be the price of the item.
After increasing 25%, it becomes 1.25x.
After decreasing 20%, it becomes x. (1.25x × 0.8 = x)
So, for every value of x, it is true.
Question 13
1 / -0

The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?

A
58

B
85

C
69

D
Either (1) or (2)

Solution

Let the digit at units place = x
Let the digit at tens place = y
Original number = 10y + x
According to the question,
x – y = 3 or y – x = 3 ……………….. (1)
Also, 10y + x + 10x + y = 143
11x + 11y = 143
x + y = 13 ……………….. (2)
Solving (1) & (2)
2x = 16
x = 8, y = 5
or x = 5 and y = 8
Required original number = 85 or 58
Hence, the correct option is (4).
Question 14
1 / -0

The sum of 100 terms of the series 1² - 2² + 3² - 4² + ……….. is

A
- 5050

B
- 4950

C
- 1101

D
240

Solution

As a² – b² = (a – b)(a + b), so 1² - 2² = (1 – 2)(1 + 2) = – 3 3² - 4² = (3 – 4)(3 + 4) = – 7
Similarly, the next terms are – 11, – 15 and so on.
So, given series = – (3 + 7 + 11 + …50 terms)
= –
= – 50 101
= – 5050
Question 15
1 / -0

If the sum of angles of a polygon is 1080°, then the number of sides of the polygon will be

A
13

B
7

C
8

D
10

Solution

1080° = 180°(n – 2)
n = 8