125876 + Interview Questions in General QA

1.	Name any two widely used cryoprotectants.
Answer» 1. Dimethyl sulphoxide 2. Glycerol

1.

Name any two widely used cryoprotectants.

Answer»

1. Dimethyl sulphoxide

2. Glycerol

2.	Point any four ways by which IPR is protected in India.
Answer» Patents, Copyrights, trade secrets and geographical indications.

Discussion

3.	Consider the following list created using HTML.D. LaptopE. DesktopF. Printer (a) START= “D” TYPE = “A” (b) START =”4″ TYPE = “A” (c) START = “4” TYPE = “I” (d) START = “D” TYPE =”l”
Answer» (d) START = “D” TYPE =”l”

Discussion

4.	Example for a sphere is ……………… A) Balls B) Laddoos C) Marbles D) All of the above
Answer» D) All of the above

Discussion

5.	Number of faces of a cube …………….. A) 6B) 8C) 12 D) 3
Answer» Correct option is A) 6

Discussion

6.	Vertices of a cube are …………….. A) 6B) 8 C) 12D) 3
Answer» Correct option is B) 8

Discussion

7.	Pyramid has …………… faces. A) 5 B) 6 C) 8 D) 12
Answer» Correct option is A) 5

Discussion

8.	Euler’s formula is ……………….A) F + E = V + 2 B) F + V = E + 2 C) F + 2 = V + E D) E + 2 = F + V
Answer» B) F + V = E + 2

Discussion

9.	Polygon means …………….. A) Many sidesB) Two sides C) One sideD) None
Answer» A) Many sides

Discussion

10.	A chord which passes through the centre of the circle is called ………………… A) Radius B) Diameter C) CircleD) Chord
Answer» Correct option is B) Diameter

Discussion

11.	Fill in the blanks.1. Three sided figure is called …………………2. Five sided figure is called ………………3. Six sided figure is called ……………….
Answer» 1. Triangle 2. Pentagon 3. Hexagon

11.

Fill in the blanks.1. Three sided figure is called …………………2. Five sided figure is called ………………3. Six sided figure is called ……………….

Answer»

1. Triangle

2. Pentagon

3. Hexagon

Discussion

12.	Four sided figure is called ………………… A) TriangleB) Pentagon C) Hexagon D) Quadrilateral
Answer» D) Quadrilateral

Discussion

13.	Many sided figure is called ……………… A) SpaceB) Polygon C) Angle D) Plane
Answer» Correct option is B) Polygon

Discussion

14.	Seven sided figure is called ………………. A) Pentagon B) Octagon C) Heptagon D) Nonagon
Answer» Correct option is C) Heptagon

Discussion

15.	A 2-dimensional closed shape formed by straight lines is called ………………. A) Plane B) SpaceC) Polygon D) None
Answer» Correct option is C) Polygon

Discussion

16.	The length of curved edge is called ……………… A) Radius B) Circle C) Circumference D) Diameter
Answer» C) Circumference circumference

Discussion

17.	Ten sided figure is known as ………………A) Decagon B) Nonagon C) SeptagonD) Hexagon
Answer» Correct option is A) Decagon

Discussion

18.	Fill in the blanks:1. Locus of the points which is at equidistant from a fixed paint is called ……………….2. The distance from the centre to any point on the circumference of circle is called ……………3. Middle point of the diameter is known as ………………
Answer» 1. Circle 2. Radius 3. Centre

18.

Fill in the blanks:1. Locus of the points which is at equidistant from a fixed paint is called ……………….2. The distance from the centre to any point on the circumference of circle is called ……………3. Middle point of the diameter is known as ………………

Answer»

1. Circle

2. Radius

3. Centre

Discussion

19.	Triangle can divides the points on the plane into …………… parts. A) 4 B) 5 C) 3 D) 2
Answer» Correct option is C) 3

Discussion

20.	Triangle has ………………. A) Three sides B) Three angles C) Three vertices D) All of these
Answer» D) All of these

Discussion

21.	Fill in the blanks:1. Another name of line of symmetry is …………………2. The shapes which have only length and breadth are known as …………………3. The 3D shape in which has length, breadth and height are equal is …………………
Answer» 1. Axis of symmetry 2. 2D shapes 3. Cube

21.

Fill in the blanks:1. Another name of line of symmetry is …………………2. The shapes which have only length and breadth are known as …………………3. The 3D shape in which has length, breadth and height are equal is …………………

Answer»

1. Axis of symmetry

2. 2D shapes

3. Cube

Discussion

22.	Fill in the blanks:1. No. of edges of square pyramid …………………2. Edges of a sphere ………………..
Answer» Correct Answer is 1. 8 2. 0 (No edges)

22.

Fill in the blanks:1. No. of edges of square pyramid …………………2. Edges of a sphere ………………..

Answer»

Correct Answer is

1. 8

2. 0 (No edges)

Discussion

23.

Match the following :Name of the polygonNumber of sides1) Cubea) 12) Pyramidb) 03) Conec) 54) Cylinderd) 75) Trianglee) 86) Heptagonf) 3

Answer»

Correct Answer is :

Name of the polygon	Number of sides
1) Cube	e) 8
2) Pyramid	c) 5
3) Cone	a) 1
4) Cylinder	b) 0
5) Triangle	f) 3
6) Heptagon	d) 7

Discussion

24.

Match the following :Number of sidesName of the polygon1) 10a) Quadrilateral2) 3b) Nonagon3) 6c) Triangle4) 9d) Decagon5) 4e) Hexagon6) 5f) Pentagon

Answer»

Correct Answer is:

Number of sides	Name of the polygon
1) 10	d) Decagon
2) 3	c) Triangle
3) 6	e) Hexagon
4) 9	b) Nonagon
5) 4	a) Quadrilateral
6) 5	f) Pentagon

Discussion

25.

Match the following :Name of the polygonNumber of sides1) Cubea) 122) Cylinderb) 83) Square pyramidc) 14) Coned) 05) Spheree) 96) Rectangular prismf) 2

Answer»

Correct Answer is:

Name of the polygon	Number of sides
1) Cube	a) 12
2) Cylinder	f) 2
3) Square pyramid	b) 8
4) Cone	c) 1
5) Sphere	d) 0
6) Rectangular prism	a) 12

Discussion

26.	Fill in the blanks:1. Example for cone shape …………………2. Example for pyramid shape ………………….3. The shape of dice is …………………
Answer» 1. Joker cap/ Heap of grains 2.Egypt Mummy 3. Cube

26.

Fill in the blanks:1. Example for cone shape …………………2. Example for pyramid shape ………………….3. The shape of dice is …………………

Answer»

1. Joker cap/ Heap of grains

2.Egypt Mummy

3. Cube

Discussion

27.	Cuboid has …………… edges. A) 6 B) 8 C) 12 D) 3
Answer» Correct option is C) 12

Discussion

28.	Give four illustrations of provision.
Answer» Four illustrations of provision are as under : Provision for depreciation, Provision for bad debt, Provision for taxtion and Provision for repairs and renewals.

28.

Give four illustrations of provision.

Answer»

Four illustrations of provision are as under :

Provision for depreciation,
Provision for bad debt,
Provision for taxtion and
Provision for repairs and renewals.

Discussion

29.

Match the following Group – AGroup – B1. Rechargea) Soil health2. Reuseb) Digudu Bavulu3. Revivec) Drip irrigation4. Reduced) Treated sewage5. Soil testinge) Checkdams and percolation tanks

Answer»

Group – A	Group – B
1. Recharge	e) Checkdams and percolation tanks
2. Reuse	d) Treated sewage
3. Revive	b) Digudu Bavulu
4. Reduce	c) Drip irrigation
5. Soil testing	a) Soil health

Discussion

30.	Solve the following systems of equations:\(\frac{2}{\sqrt x}\) + \(\frac{3}{\sqrt y}\) = 2\(\frac{4}{\sqrt x}\) - \(\frac{9}{\sqrt y}\) = - 1
Answer» \(\frac{2}{\sqrt x}\) + \(\frac{3}{\sqrt y}\) = 2 \(\frac{4}{\sqrt x}\) - \(\frac{9}{\sqrt y}\) = - 1 Multiplying eq1 by 3 and adding to eq2 ⇒ 10/√x = 5 ⇒ √x = 2 ⇒ x = 4 Thus, ⇒ 2/√4 + 3/√y = 2 ⇒ y = 9

30.

Solve the following systems of equations:\(\frac{2}{\sqrt x}\) + \(\frac{3}{\sqrt y}\) = 2\(\frac{4}{\sqrt x}\) - \(\frac{9}{\sqrt y}\) = - 1

Answer»

\(\frac{2}{\sqrt x}\) + \(\frac{3}{\sqrt y}\) = 2

\(\frac{4}{\sqrt x}\) - \(\frac{9}{\sqrt y}\) = - 1

Multiplying eq1 by 3 and adding to eq2

⇒ 10/√x = 5

⇒ √x = 2

⇒ x = 4

Thus,

⇒ 2/√4 + 3/√y = 2

⇒ y = 9

Discussion

31.	Solve the following system of equations:2/x + 3/y = 9/xy 4/x + 9/y = 21/xy
Answer» Taking LCM for both the given equations, we have (2y + 3x)/ xy = 9/xy ⇒ 3x + 2y = 9………. (i) (4y + 9x)/ xy = 21/xy ⇒ 9x + 4y = 21………(ii) Performing (ii) – (i) x 2 ⇒ 9x + 4y – 2(3x + 2y) = 21 – (9×2) ⇒ 3x = 3 ⇒ x = 1 Using x = 1 in (i), we find y 3(1) + 2y = 9 ⇒ y = \(\frac{6}{2}\) ⇒ y = 3 Thus, the solution for the given set of equations is x = 1 and y = 3.

31.

Solve the following system of equations:2/x + 3/y = 9/xy 4/x + 9/y = 21/xy

Answer»

Taking LCM for both the given equations, we have

(2y + 3x)/ xy = 9/xy

⇒ 3x + 2y = 9………. (i)

(4y + 9x)/ xy = 21/xy

⇒ 9x + 4y = 21………(ii)

Performing (ii) – (i) x 2

⇒ 9x + 4y – 2(3x + 2y) = 21 – (9×2)

⇒ 3x = 3

⇒ x = 1

Using x = 1 in (i), we find y

3(1) + 2y = 9

⇒ y = \(\frac{6}{2}\)

⇒ y = 3

Thus, the solution for the given set of equations is x = 1 and y = 3.

Discussion

32.	We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Answer» i) Given: 2x + 3y – 8 = 0 The lines are intersecting lines. Let the other linear equation be ax + by + c = 0 ∴ a₁/a₂ ≠ b₁/b₂; we have to choose appropriate values satisfying the condition above. Thus the other equation may be 3x + 5y – 6 =0 ii) Parallel line a₁/a₂ = b₁/b₂ ≠ c₁/c₂ ⇒ 2x + 3y – 8 = 0 ⇒ 4x + 6y – 10 = 0 iii) Coincident lines a1/a2 = b1/b2 = c1/c2 ⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0

32.

We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.

Answer»

i) Given: 2x + 3y – 8 = 0

The lines are intersecting lines.

Let the other linear equation be ax + by + c = 0

∴ a₁/a₂ ≠ b₁/b₂; we have to choose appropriate values satisfying the condition above.

Thus the other equation may be 3x + 5y – 6 =0

ii) Parallel line a₁/a₂ = b₁/b₂ ≠ c₁/c₂

⇒ 2x + 3y – 8 = 0

⇒ 4x + 6y – 10 = 0

iii) Coincident lines a1/a2 = b1/b2 = c1/c2

⇒ 2x + 3y – 8 = 0

⇒ 8x + 12y – 32 = 0

Discussion

33.	List the prime numbers from 25 to 100 and say how many they are.
Answer» 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. There are 16 prime numbers from 25 to 100.

33.

List the prime numbers from 25 to 100 and say how many they are.

Answer»

29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

There are 16 prime numbers from 25 to 100.

Discussion

34.	Express the following numbers as the sum of two odd primes.(i) 18(ii) 24(iii) 36(iv) 44
Answer» i) 18 = 7 + 11 = 5 + 13 ii) 24 = 5 + 19 = 7 + 17 = 11 + 13 iii) 36 = 5 + 31 = 7 + 29 = 13 + 23 = 47 + 19 iv) 44 = 3 + 41 = 7 + 37 = 13 + 31

34.

Express the following numbers as the sum of two odd primes.(i) 18(ii) 24(iii) 36(iv) 44

Answer»

i) 18 = 7 + 11 = 5 + 13

ii) 24 = 5 + 19 = 7 + 17 = 11 + 13

iii) 36 = 5 + 31 = 7 + 29 = 13 + 23 = 47 + 19

iv) 44 = 3 + 41 = 7 + 37 = 13 + 31

Discussion

35.	From the following numbers identify different pairs of co-primes. 2, 3,4, 5,6,7, 8, 9 and 10.
Answer» The numbers which have only 1 as the common factor are called co-primes, (or) Numbers having no common factors, other than 1 are called co-primes. 2,3; 2, 5; 2, 7; 2, 9; 3,4; 3, 5; 3, 7; 3, 8; 3,10; 4,5; 4, 7; 4, 9; 5, 6; 5, 7; 5,8; 5, 9; 6, 7; 8, 9 and 9,10. These are the different pairs of co-primes with 2, 3, 4, 5, 6, 7, 8, 9 and 10. 1) Any two primes always forms a pair of co-primes. 2) Any two consecutive numbers always form a pair of co-primes. 3) Any two primes cilways form a pair of co-primes. .

35.

From the following numbers identify different pairs of co-primes. 2, 3,4, 5,6,7, 8, 9 and 10.

Answer»

The numbers which have only 1 as the common factor are called co-primes,

(or)

Numbers having no common factors, other than 1 are called co-primes.

2,3; 2, 5; 2, 7; 2, 9; 3,4; 3, 5; 3, 7; 3, 8; 3,10; 4,5; 4, 7; 4, 9; 5, 6; 5, 7; 5,8; 5, 9; 6, 7; 8, 9 and 9,10.

These are the different pairs of co-primes with 2, 3, 4, 5, 6, 7, 8, 9 and 10.

1) Any two primes always forms a pair of co-primes.

2) Any two consecutive numbers always form a pair of co-primes.

3) Any two primes cilways form a pair of co-primes. .

Discussion

36.	Which of the following are pairs of co-primes? i. 8, 14 ii. 4, 5 iii. 17, 19 iv. 27, 15
Answer» i. Factors of 8: 1, 2, 4, 8 Factors of 14: 1, 2, 7, 14 ∴ Common factors of 8 and 14: 1, 2 ∴ 8 and 14 are not a pair of co-prime numbers. ii. Factors of 4: 1, 4 Factors of 5 : 1, 5 ∴ Common factors of 4 and 5 : 1 ∴ 4 and 5 are a pair of co-prime numbers. iii. Factors of 17 : 1, 17 Factors of 19 : 1, 19 ∴ Common factors of 17 and 19 : 1 ∴ 17 and 19 are a pair of co-prime numbers. iv. Factors of 27: 1, 3, 9, 27 Factors of 15: 1, 3, 5, 15. ∴ Common factors of 27 and 15 : 1, 3 ∴ 27 and 15 are not a pair of co-prime numbers.

36.

Which of the following are pairs of co-primes? i. 8, 14 ii. 4, 5 iii. 17, 19 iv. 27, 15

Answer»

i. Factors of 8: 1, 2, 4, 8

Factors of 14: 1, 2, 7, 14

∴ Common factors of 8 and 14: 1, 2

∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4

Factors of 5 : 1, 5

∴ Common factors of 4 and 5 : 1

∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17 : 1, 17

Factors of 19 : 1, 19

∴ Common factors of 17 and 19 : 1

∴ 17 and 19 are a pair of co-prime numbers.

iv. Factors of 27: 1, 3, 9, 27

Factors of 15: 1, 3, 5, 15.

∴ Common factors of 27 and 15 : 1, 3

∴ 27 and 15 are not a pair of co-prime numbers.

Discussion

37.	In the adjoining figure ABCD, the angles x and y are (A) 600 , 300 (B) 300 , 600 (C) 450 , 450 (D) 900 , 900
Answer» The correct option is (A).

Discussion

38.	In a parallelogram the sum of the angle bisector of two adjacent angles is (A) 300 (B) 450 (C) 600 (D) 900
Answer» The correct option is (D).

Discussion

39.	In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of parallelogram ABCD.
Answer» Construct AM ⊥ DC and CL ⊥ AB. Extend the lines DC and AB and join AC. We know that Area of the quadrilateral ABCD = area of triangle ABD + area of triangle DCB From the figure we know that Area of △ ABD = Area of △ DCB So it can be written as Area of the quadrilateral ABCD = 2 (Area of △ ABD) We get Area of △ ABD = ½ (area of quadrilateral ABCD) ……… (1) Area of quadrilateral ABCD = area of △ ABC + area of △ CDA From the figure we know that Area of △ ABC = Area of △ CDA So it can be written as Area of the quadrilateral ABCD = 2 (Area of △ ABC) We get Area of △ ABC = ½ (area of quadrilateral ABCD) ……… (2) Using the equations (1) and (2) Area of △ ABD = Area of △ ABC = ½ BD = ½ CL It can be written as CL = BD In the same way we get DC = AB and AD = BC Hence, ABCD is a parallelogram Area of parallelogram ABCD = base × height = 5 × 7 = 35 cm² Therefore, area of parallelogram ABCD is 35 cm².

39.

In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of parallelogram ABCD.

Answer»

Construct AM ⊥ DC and CL ⊥ AB. Extend the lines DC and AB and join AC.

We know that

Area of the quadrilateral ABCD = area of triangle ABD + area of triangle DCB

From the figure we know that

Area of △ ABD = Area of △ DCB

So it can be written as

Area of the quadrilateral ABCD = 2 (Area of △ ABD)

We get

Area of △ ABD = ½ (area of quadrilateral ABCD) ……… (1)

Area of quadrilateral ABCD = area of △ ABC + area of △ CDA

From the figure we know that

Area of △ ABC = Area of △ CDA

So it can be written as

Area of the quadrilateral ABCD = 2 (Area of △ ABC)

We get

Area of △ ABC = ½ (area of quadrilateral ABCD) ……… (2)

Using the equations (1) and (2)

Area of △ ABD = Area of △ ABC = ½ BD = ½ CL

It can be written as

CL = BD

In the same way we get

DC = AB and AD = BC

Hence, ABCD is a parallelogram

Area of parallelogram ABCD = base × height

= 5 × 7

= 35 cm²

Therefore, area of parallelogram ABCD is 35 cm².

Discussion

40.	P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (PAB) = ar (BQC).
Answer» Data: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. To Prove: ar(∆APB) = ar(∆BQC) Proof: ABCD is a parallelogram. ∴ AB \|\| DC AB = DC AD \|\| BC AD = BC Now ∆APB and ABCD are on same base AB and in between AB \|\| DC ∴ Area(∆APB) = Area ABCD) ……… (i) Similarly, ∆BQC and BADC are on the same base BC and in between BC \|\| AD. ∴ Area(∆BQC) = Area( ABCD) ………. (ii) From (i) and (ii) ∴ Area(∆APB) = Area (∆BQC).

40.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (PAB) = ar (BQC).

Answer»

Data: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

To Prove: ar(∆APB) = ar(∆BQC)

Proof: ABCD is a parallelogram.

∴ AB || DC

AB = DC

AD || BC

AD = BC

Now ∆APB and ABCD are on same base AB and in between AB || DC

∴ Area(∆APB) = Area ABCD) ……… (i)

Similarly, ∆BQC and BADC are on the same base

BC and in between BC || AD.

∴ Area(∆BQC) = Area( ABCD) ………. (ii)

From (i) and (ii)

∴ Area(∆APB) = Area (∆BQC).

Discussion

41.	Check which of the following are solutions of the equation x − 2y = 4 and which are not:(i) ( √2 , 4√2 )(ii) ( 1, 1)
Answer» (i) ( √2 , 4√2 ) Putting x = √2 and y = 4√2 in the L.H.S of the given equation, x - 2y = √2 - 2(4√2) = √2 - 8√2 = -7√2 ≠ 4 L.H.S ≠ R.H.S Therefore,(√2, 4√2) is not a solution of this equation. (ii) (1, 1) Putting x = 1 and y = 1 in the L.H.S of the given equation, x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4 L.H.S ≠ R.H.S Therefore, (1, 1) is not a solution of this equation.

41.

Check which of the following are solutions of the equation x − 2y = 4 and which are not:(i) ( √2 , 4√2 )(ii) ( 1, 1)

Answer»

(i) ( √2 , 4√2 )

Putting x = √2 and y = 4√2 in the L.H.S of the given equation,

x - 2y = √2 - 2(4√2)

= √2 - 8√2 = -7√2 ≠ 4

L.H.S ≠ R.H.S

Therefore,(√2, 4√2) is not a solution of this equation.

(ii) (1, 1)

Putting x = 1 and y = 1 in the L.H.S of the given equation,

x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4 L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

Discussion

42.	Check which of the following are solutions of the equation x − 2y = 4 and which are not:(i) ( 0, 2)(ii) (2, 0)(iii) (4, 0)
Answer» (i) (0, 2) Putting x = 0 and y = 2 in the L.H.S of the given equation, x − 2y = 0 − 2×2 = − 4 ≠ 4 L.H.S ≠ R.H.S Therefore, (0, 2) is not a solution of this equation. (ii) (2, 0) Putting x = 2 and y = 0 in the L.H.S of the given equation, x − 2y = 2 − 2 × 0 = 2 ≠ 4 L.H.S ≠ R.H.S Therefore, (2, 0) is not a solution of this equation. (iii) (4, 0) Putting x = 4 and y = 0 in the L.H.S of the given equation, x − 2y = 4 − 2(0) = 4 = R.H.S Therefore, (4, 0) is a solution of this equation.

42.

Check which of the following are solutions of the equation x − 2y = 4 and which are not:(i) ( 0, 2)(ii) (2, 0)(iii) (4, 0)

Answer»

(i) (0, 2)

Putting x = 0 and y = 2 in the L.H.S of the given equation,

x − 2y = 0 − 2×2 = − 4 ≠ 4

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution of this equation.

(ii) (2, 0)

Putting x = 2 and y = 0 in the L.H.S of the given equation,

x − 2y = 2 − 2 × 0 = 2 ≠ 4

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution of this equation.

(iii) (4, 0)

Putting x = 4 and y = 0 in the L.H.S of the given equation,

x − 2y = 4 − 2(0)

= 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

Discussion

43.	Write four solutions for each of the following equations:x = 4y
Answer» x = 4y For x = 0, 0 = 4y ⇒ y = 0 x = 4(1) = 4 x = 4(−1) ⇒ x = −4 Therefore, (−4, −1) is a solution of this equation. For x = 2, 2 = 4y ⇒ y = 2/4 = 1/2 Therefore, (2,1/2) is a solution of this equation.

Discussion

44.	Find the value of x so that (–2)3 × (–2)–6 = (–2)2x–1.
Answer» From the laws of exponents , we know that, a^m × aⁿ = a^(m+n) So , we have , (–2)³ × (–2)^–6 =(-2)^(3)+(-6) = (-2)^(-3) Now , we have , (-2)^(-3)= (-2)^(2x-1) On comparing both the sides , we get -3 = 2x-1 2x = -3 + 1 2x = -2 x = -2/2 = -1 ∴ x = -1

44.

Find the value of x so that (–2)3 × (–2)–6 = (–2)2x–1.

Answer»

From the laws of exponents , we know that,

a^m × aⁿ = a^(m+n)

So , we have , (–2)³ × (–2)^–6

=(-2)^(3)+(-6)

= (-2)^(-3)

Now , we have , (-2)^(-3)= (-2)^(2x-1)

On comparing both the sides , we get

-3 = 2x-1

2x = -3 + 1

2x = -2

x = -2/2 = -1

∴ x = -1

Discussion

45.	7x+3 = 5x+3 then x = ……………….A) – 1 B) – 3C) – 4 D) 5
Answer» Correct option is (B) –3 \(7^{x+3}=5^{x+3}\) This happens only when x+3 = 0 \((\because7^0=1=5^0)\) \(\Rightarrow\) x = -3 Correct option is B) – 3

45.

7x+3 = 5x+3 then x = ……………….A) – 1 B) – 3C) – 4 D) 5

Answer»

Correct option is (B) –3

\(7^{x+3}=5^{x+3}\)

This happens only when x+3 = 0 \((\because7^0=1=5^0)\)

\(\Rightarrow\) x = -3

Correct option is B) – 3

Discussion

46.	Fill in the blanks to make the statements true.[(2/13)-6 ÷ (2/13)3]3 × (2/13)-9 = ______
Answer» [(2/13)^-6÷(2/13)³]³ × (2/13)^-9 = (2/13)^-36 Explanation: [(2/13)^-6÷(2/13)³]³ × (2/13)^-9 = [(2/13)^-6-3]³ × (2/13)^-9 = [(2/13)^-9]³ × (2/13)^-9 = (2/13)^-9×3 × (2/13)^-9 = (2/13)^-27 × (2/13)^-9 = (2/13)^-27-9 = (2/13)^-36

46.

Fill in the blanks to make the statements true.[(2/13)-6 ÷ (2/13)3]3 × (2/13)-9 = ______

Answer»

[(2/13)^-6÷(2/13)³]³ × (2/13)^-9 = (2/13)^-36

Explanation:

[(2/13)^-6÷(2/13)³]³ × (2/13)^-9

= [(2/13)^-6-3]³ × (2/13)^-9

= [(2/13)^-9]³ × (2/13)^-9

= (2/13)^-9×3 × (2/13)^-9

= (2/13)^-27 × (2/13)^-9

= (2/13)^-27-9

= (2/13)^-36

Discussion

47.

Fill in the blanks to make the statements true.a3 × a–10 = __________.

Answer»

We know the formula –

⇒ a^m× aⁿ = a^m+n

⇒ a³× a^-10 =a^{3 – 10}

∴ a³× a^-10 =a^{– 7}

Discussion

48.	What is the angle of vision?
Answer» The maximum angle, at which we are able to see the whole object is called the angle of vision.

Discussion

49.

Match the adjectives in column ‘A’ with the nouns in column ‘B’: 'A' 'B' (1) new (a) prayer (2) silent (b) overcoat (3) first (c) expression (4) peculiar (d) flight

Answer»

'A'	'B'
(1) new	(b) overcoat
(2) silent	(a) prayer
(3) first	(d) flight
(4) peculiar	(c) expression

Discussion

50.

Give scientific reason :We are able to understand the smell of the first showers of rain or the sudden changes in the climate.

Answer»

1. We are able to understand any smell because of our olfactory mucosa and olfactory lobes of the brain.

2. Volatile substances are received by the olfactoreceptors in the nose.

3. Nerve impulse generated are carried by olfactory nerve and transmitted to brain where the impulse is interpreted.

4. The characteristic earthy smell is due to a compound ‘geosmin’.

5. Geosmin is produced by some species of Streptomyces [ gram positive soil bacterium],

6. Similarly, sudden change in the climate is easily noticed in the form of temperature change in the surrounding.

7. This change is detected by caloreceptors of the skin. From these receptors the signal is transmitted to CNS where the change is perceived.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

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7x+3 = 5x+3 then x = ……………….A) – 1 B) – 3C) – 4 D) 5

Fill in the blanks to make the statements true.[(2/13)-6 ÷ (2/13)3]3 × (2/13)-9 = ______

Fill in the blanks to make the statements true.a3 × a–10 = __________.

What is the angle of vision?

Match the adjectives in column ‘A’ with the nouns in column ‘B’: 'A' 'B' (1) new (a) prayer (2) silent (b) overcoat (3) first (c) expression (4) peculiar (d) flight

Give scientific reason :We are able to understand the smell of the first showers of rain or the sudden changes in the climate.