125876 + Interview Questions in General QA

1.	What does the law do to unjust customs?
Answer» The laws abolish unjust customs.

Discussion

2.	There are several ………………….. customs and traditions in our society. (a) good (b) bad (c) ugly
Answer» Correct option is (a) good

Discussion

3.	What are the differences between “Logical error” and “Syntax error”?
Answer» Logical Error : Logical errors occur when there is an incorrect usage of variable / operator / order of execution etc. It is also called as Semantic Error. Syntax Error : Syntax errors occur when grammatical rules of C++ are violated.

3.

What are the differences between “Logical error” and “Syntax error”?

Answer»

Logical Error : Logical errors occur when there is an incorrect usage of variable / operator / order of execution etc. It is also called as Semantic Error.

Syntax Error : Syntax errors occur when grammatical rules of C++ are violated.

Discussion

4.	In Fig. ABCD is a trapezium in which AB\|\|DC. Prove that ar(Δ AOD) = ar(Δ BOC).
Answer» Given that, ABCD is a trapezium with AB ‖ DC To prove : Area (ΔAOD) = Area (ΔBOC) Proof : Since, ΔABC and ΔABD are on the same base AB and between the same parallels AB and DC Therefore, Area (Δ ABC) = Area (Δ ABD) Area (Δ ABC) – Area (Δ AOB) = Area (ΔABD) – Area (Δ AOB) Area (Δ AOD) = Area (Δ BOC) Hence, proved

4.

In Fig. ABCD is a trapezium in which AB||DC. Prove that ar(Δ AOD) = ar(Δ BOC).

Answer»

Given that,

ABCD is a trapezium with AB ‖ DC

To prove :

Area (ΔAOD) = Area (ΔBOC)

Proof :

Since,

ΔABC and ΔABD are on the same base AB and between the same parallels AB and DC

Therefore,

Area (Δ ABC) = Area (Δ ABD)

Area (Δ ABC) – Area (Δ AOB)

= Area (ΔABD) – Area (Δ AOB)

Area (Δ AOD) = Area (Δ BOC)

Hence, proved

Discussion

5.	In Fig. X and Y are the mid-point of AC and AB respectively, QP\|\|BC and CYQ and BXP are straight lines. Prove that : ar(Δ ABP) = ar(Δ ACQ).
Answer» In a ΔAXP and ΔCXB, ∠PAX = XCB (Alternative angles AP \|\| BC) AX = CX (Given) ∠AXP = ∠CXB (Vertically opposite angles) ΔAXP ≅ ΔCXB (By ASA rule) AP = BC (By c.p.c.t) ...(i) Similarly, QA = BC ...(ii) From (i) and (ii), we get AP = QA Now, AP \|\| BC And, AP = QA Area (ΔAPB) = Area (ΔACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)

5.

In Fig. X and Y are the mid-point of AC and AB respectively, QP||BC and CYQ and BXP are straight lines. Prove that : ar(Δ ABP) = ar(Δ ACQ).

Answer»

In a ΔAXP and ΔCXB,

∠PAX = XCB

(Alternative angles AP || BC)

AX = CX (Given)

∠AXP = ∠CXB

(Vertically opposite angles)

ΔAXP ≅ ΔCXB (By ASA rule)

AP = BC (By c.p.c.t) ...(i)

Similarly,

QA = BC ...(ii)

From (i) and (ii), we get

AP = QA

Now,

AP || BC

And,

AP = QA

Area (ΔAPB) = Area (ΔACQ)

(Therefore, Triangles having equal bases and between the same parallels QP and BC)

Discussion

6.	In a Δ ABC, if L and M are points on AB and AC respectively such that LM\|\|BC. Prove that : (i) ar(ΔLCM) = ar(ΔLBM) (ii) ar(ΔLBC) = ar(ΔMBC) (iii) ar(ΔABM) = ar(ΔACL) (iv) ar(ΔLOB) = ar(ΔMOC)
Answer» (i) Clearly, Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC. Therefore, Area (ΔLMB) = Area (ΔLMC) ...(i) (ii) We observe that, Triangles LBC and MBC are on the same base BC and between the same parallels LM and BC. Therefore, Area (ΔLBC) = Area (ΔMBC) ...(ii) (iii) We have, Area (ΔLMB) = Area (ΔLMC) [From (i)] Area (ΔALM) + Area (ΔLMB) = Area (ΔALM) = Area (ΔLMC) Area (ΔABM) = Area (ΔACL) (iv) We have, Area (ΔLBC) = Area (ΔMBC) [From (ii)] Area (ΔLBC) - Area (ΔBOC) = Area (ΔMBC) - Area (ΔBOC) Area (ΔLOB) =Area (ΔMOC)

6.

In a Δ ABC, if L and M are points on AB and AC respectively such that LM||BC. Prove that : (i) ar(ΔLCM) = ar(ΔLBM) (ii) ar(ΔLBC) = ar(ΔMBC) (iii) ar(ΔABM) = ar(ΔACL) (iv) ar(ΔLOB) = ar(ΔMOC)

Answer»

(i) Clearly,

Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

Therefore,

Area (ΔLMB) = Area (ΔLMC) ...(i)

(ii) We observe that,

Triangles LBC and MBC are on the same base BC and between the same parallels LM and BC.

Therefore,

Area (ΔLBC) = Area (ΔMBC) ...(ii)

(iii) We have,

Area (ΔLMB) = Area (ΔLMC) [From (i)]

Area (ΔALM) + Area (ΔLMB) = Area (ΔALM) = Area (ΔLMC)

Area (ΔABM) = Area (ΔACL)

(iv) We have,

Area (ΔLBC) = Area (ΔMBC) [From (ii)]

Area (ΔLBC) - Area (ΔBOC) = Area (ΔMBC) - Area (ΔBOC)

Area (ΔLOB) =Area (ΔMOC)

Discussion

7.	In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP. Prove that : (i) ar(ΔPBQ) = ar(ΔARC) (ii) ar(ΔPQR) = \(\frac{1}{2}\)ar(ΔARC) (iii) ar(ΔRQC) = \(\frac{3}{8}\)ar(ΔABC)
Answer» (i) We know that, Each median of a triangle divides it into two triangles of equal area. Since, CR is a median of ΔCAP Therefore, Area (ΔCRA) = Area (ΔCAP) ...(i) Also, CP is a median of ΔCAB Therefore, Area (ΔCAP) = Area (ΔCPB) ...(ii) From (i) and (ii), we get Therefore, Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCPB) ...(iii) PQ is a median of ΔPBC Therefore, Area (ΔCPB) = 2 Area (ΔPQB) ...(iv) From (iii) and (iv), we get Area (ΔARC) = Area (ΔPBQ) ...(v) (ii) Since, QP and QR medians of ΔQAB and QAP respectively. Area (ΔQAP) = Area (ΔQBP) ...(vi) And, Area (ΔQAP) = 2 Area (ΔQRP) ...(vii) From (vi) and (vii), we get Area (ΔPRQ) = Area (ΔPBQ)...(viii) From (v) and (viii), we get Area (ΔPRQ) = Area (ΔARC) ...(ix) (iii) Since, CR is a median of ΔCAP Therefore, Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCAP) = \(\frac{1}{2}\)x \(\frac{1}{2}\) Area (ΔABC) (Therefore, CP is a median of Δ ABC) =\(\frac{1}{4}\) Area (ΔABC) ...(x) Since, RQ is a median of Δ RBC. Therefore, Area (ΔRQC) = \(\frac{1}{2}\)Area (ΔRBC) = \(\frac{1}{2}\)[Area (ΔABC) – Area (ΔARC)] = \(\frac{1}{2}\)[Area (ΔABC) - Area (ΔABC)] = \(\frac{3}{4}\)Area (ΔABC)

7.

In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP. Prove that : (i) ar(ΔPBQ) = ar(ΔARC) (ii) ar(ΔPQR) = \(\frac{1}{2}\)ar(ΔARC) (iii) ar(ΔRQC) = \(\frac{3}{8}\)ar(ΔABC)

Answer»

(i) We know that,

Each median of a triangle divides it into two triangles of equal area.

Since,

CR is a median of ΔCAP

Therefore,

Area (ΔCRA) = Area (ΔCAP) ...(i)

Also,

CP is a median of ΔCAB

Therefore,

Area (ΔCAP) = Area (ΔCPB) ...(ii)

From (i) and (ii), we get

Therefore,

Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCPB) ...(iii)

PQ is a median of ΔPBC

Therefore,

Area (ΔCPB) = 2 Area (ΔPQB) ...(iv)

From (iii) and (iv), we get

Area (ΔARC) = Area (ΔPBQ) ...(v)

(ii) Since,

QP and QR medians of ΔQAB and QAP respectively.

Area (ΔQAP) = Area (ΔQBP) ...(vi)

And,

Area (ΔQAP) = 2 Area (ΔQRP) ...(vii)

From (vi) and (vii), we get

Area (ΔPRQ) = Area (ΔPBQ)...(viii)

From (v) and (viii), we get

Area (ΔPRQ) = Area (ΔARC) ...(ix)

(iii) Since,

CR is a median of ΔCAP

Therefore,

Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCAP)

= \(\frac{1}{2}\)x \(\frac{1}{2}\) Area (ΔABC)

(Therefore, CP is a median of Δ ABC)

=\(\frac{1}{4}\) Area (ΔABC) ...(x)

Since,

RQ is a median of Δ RBC.

Therefore,

Area (ΔRQC) = \(\frac{1}{2}\)Area (ΔRBC)

= \(\frac{1}{2}\)[Area (ΔABC) – Area (ΔARC)]

= \(\frac{1}{2}\)[Area (ΔABC) - Area (ΔABC)]

= \(\frac{3}{4}\)Area (ΔABC)

Discussion

8.	D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, Prove that, ar(ΔBOE) = \(\frac{1}{8}\)ar(ΔABC)
Answer» Join A and D to get AD median (Median divides the triangle into two triangles of equal area) Therefore, Area(ΔABD) = \(\frac{1}{2}\)Area (ΔABC) Now, Join A and E to get AE median Similarly, We can prove that, Area(ΔABE) = \(\frac{1}{2}\)Area (ΔABD) Area(ΔABE) = \(\frac{1}{4}\)ABC (Area (ΔABD) = \(\frac{1}{2}\)Area (ΔABC) (i) Join B and O and we get BO median Now, Area(ΔBOE) = \(\frac{1}{2}\)Area (ΔABE) Area(ΔBOE) = \(\frac{1}{2}\) x \(\frac{1}{4}\)Area (ΔABC) Area(ΔBOE) = \(\frac{1}{8}\)Area (ΔABC)

8.

D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, Prove that, ar(ΔBOE) = \(\frac{1}{8}\)ar(ΔABC)

Answer»

Join A and D to get AD median

(Median divides the triangle into two triangles of equal area)

Therefore,

Area(ΔABD) = \(\frac{1}{2}\)Area (ΔABC)

Now,

Join A and E to get AE median

Similarly,

We can prove that,

Area(ΔABE) = \(\frac{1}{2}\)Area (ΔABD)

Area(ΔABE) = \(\frac{1}{4}\)ABC (Area (ΔABD) = \(\frac{1}{2}\)Area (ΔABC) (i)

Join B and O and we get BO median

Now,

Area(ΔBOE) = \(\frac{1}{2}\)Area (ΔABE)

Area(ΔBOE) = \(\frac{1}{2}\) x \(\frac{1}{4}\)Area (ΔABC)

Area(ΔBOE) = \(\frac{1}{8}\)Area (ΔABC)

Discussion

9.	In Fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Answer» Draw AL perpendicular to DC And, BM perpendicular DC Then, AL = BM = 4 cm And, LM = 7 cm In Δ ADL, we have AD² = AL²⁺ DL² 25 = 16 + DL² DL = 3 cm Similarly, MC = \(\sqrt{BC^2-BM^2}\) = \(\sqrt{25-16}\) = 3 cm Therefore, x = CD = CM + ML + LD = 3 + 7 + 3 = 13 cm Area of trapezium ABCD = \(\frac{1}{2}\)(AB + CD) x AL = \(\frac{1}{2}\)(7 + 13) x 4 = 40 cm²

9.

In Fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Answer»

Draw AL perpendicular to DC

And,

BM perpendicular DC

Then,

AL = BM = 4 cm

And,

LM = 7 cm

In Δ ADL, we have

AD² = AL²⁺ DL²

25 = 16 + DL²

DL = 3 cm

Similarly,

MC = \(\sqrt{BC^2-BM^2}\)

= \(\sqrt{25-16}\)

= 3 cm

Therefore,

x = CD

= CM + ML + LD

= 3 + 7 + 3

= 13 cm

Area of trapezium ABCD = \(\frac{1}{2}\)(AB + CD) x AL

= \(\frac{1}{2}\)(7 + 13) x 4

= 40 cm²

Discussion

10.	Compute the area of trapezium PQRS in Fig.
Answer» We have, Area of trapezium PQRS = Area of rectangle PSRT + Area (Δ QRT) Area of trapezium PQRS = PT x RT + (QT x RT) = 8 x RT + (8 x RT) = 12 x RT In Δ QRT, we have QR² = QT² + RT² RT² = QR² - QT² RT² = (17)² - (8)² = 225 = 15 Hence, Area of trapezium PQRS = 12 x 15 = 180 cm²

10.

Compute the area of trapezium PQRS in Fig.

Answer»

We have,

Area of trapezium PQRS = Area of rectangle PSRT + Area (Δ QRT)

Area of trapezium PQRS = PT x RT + (QT x RT) = 8 x RT + (8 x RT) = 12 x RT

In Δ QRT, we have

QR² = QT² + RT²

RT² = QR² - QT²

RT² = (17)² - (8)²

= 225

= 15

Hence,

Area of trapezium PQRS = 12 x 15

= 180 cm²

Discussion

11.	In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.
Answer» Since, The mid-point of the hypotenuse of a right triangle is equidistant from the vertices Therefore, CA = CB = OC CA = CB = 6.5 cm AB = 13 cm In right (ΔOAB) We have, AB² = OB² - OA² 13² = OB² + 12² OB = 5 cm Therefore, Area (Δ AOB) = \(\frac{1}{2}\)(OA x OB) =\(\frac{1}{2}\)(12 x 5) = 30 cm²

11.

In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

Answer»

Since,

The mid-point of the hypotenuse of a right triangle is equidistant from the vertices

Therefore,

CA = CB = OC

CA = CB = 6.5 cm

AB = 13 cm

In right (ΔOAB)

We have,

AB² = OB² - OA²

13² = OB² + 12²

OB = 5 cm

Therefore,

Area (Δ AOB) = \(\frac{1}{2}\)(OA x OB)

=\(\frac{1}{2}\)(12 x 5)

= 30 cm²

Discussion

12.	In figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.
Answer» Given: A triangle AOB, with ∠AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm As we know, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. So, CB = CA = OC = 6.5 cm AB = 2 CB = 2 x 6.5 cm = 13 cm In right ΔOAB: Using Pythagorean Theorem, we get AB² = OB² + OA² 13² = OB² + 12² OB² = 169 – 144 = 25 or OB = 5 cm Now, Area of ΔAOB = 1/2(Base x height) cm² = 1/2(12 x 5) cm² = 30 cm²

12.

In figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.

Answer»

Given: A triangle AOB, with ∠AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm

As we know, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.

So, CB = CA = OC = 6.5 cm

AB = 2 CB = 2 x 6.5 cm = 13 cm

In right ΔOAB:

Using Pythagorean Theorem, we get

AB² = OB² + OA²

13² = OB² + 12²

OB² = 169 – 144 = 25

or OB = 5 cm

Now, Area of ΔAOB = 1/2(Base x height) cm²

= 1/2(12 x 5) cm²

= 30 cm²

Discussion

13.	Simplify and write the result in decimal form :(1÷ 2/9) + (1 ÷ 3 1/5) + (1 ÷ 2 2/3)
Answer» Correct answer is 5.1875

Discussion

14.	Fill in the blanks to make the statement true.To divide a decimal number by 100, we shift the decimal point in the number to the ________ by ______ places.
Answer» To divide a decimal number by 100, we shift the decimal point in the number to the left by two places.

Discussion

15.	Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer» It can be observed that point (2, 14) satisfies the equation 7x − y = 0 and x − y + 12 = 0. Therefore, 7x − y = 0 and x − y + 12 = 0 are two lines passing through point (2, 14). As it is known that through one point, infinite number of lines can pass through, therefore, there are infinite lines of such type passing through the given point.

15.

Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer»

It can be observed that point (2, 14) satisfies the equation 7x − y = 0 and x − y + 12 = 0.

Therefore, 7x − y = 0 and x − y + 12 = 0 are two lines passing through point (2, 14).

As it is known that through one point, infinite number of lines can pass through, therefore, there are infinite lines of such type passing through the given point.

Discussion

16.	To find the distance around a circular disc, multiply the diameter of the disc by 3.14. What is the distance around the disc when :(a) the diameter is 18.7 cm?(b) the radius is 6.45 cm?
Answer» Correct answer is (a) 58.718 cm (b) 40.506 cm

16.

To find the distance around a circular disc, multiply the diameter of the disc by 3.14. What is the distance around the disc when :(a) the diameter is 18.7 cm?(b) the radius is 6.45 cm?

Answer»

Correct answer is

(a) 58.718 cm (b) 40.506 cm

Discussion

17.

Some balls have been assigned a tag with decimal numbers. The balls are to be placed on the basis of the following :• All balls with tag number less than 1/8 should be put in box 1.• All balls with tag number between 3/8 and 5/8 should be put in box 2.• All balls with tag number more than 7/8 should be put in box 3.Put the balls in appropriate number of box by arrow.

Answer»

Box 1	Box 2	Box 3
0.096	0.376	1.808
0.001	0.4200	0.987
0.066	0.62	11.00
0.0864	0.578	0.888

Discussion

18.	What is the Error in the question. A student compared – 1/4 and –0.3. He changed –1/ 4 to the decimal –0.25 and wrote, “Since 0.3 is greater than 0.25, –0.3 is greater than –0.25”. What was the student’s error?
Answer» Error –0.30 > –0.25

Discussion

19.	A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by 1.414. Find the length of the diagonal when :(a) The length of a side of the square is 8.3 cm.(b) The length of a side of the square is exactly 7.875 cm.
Answer» Correct answer is (a) 11.74 cm (approxmatlly) (b) 11.14 cm (approxmatlly)

Discussion

20.	State whether the statement is True or False. A reciprocal of a fraction is obtained by inverting it upside down.
Answer» A reciprocal of a fraction is obtained by inverting it upside down. True

20.

State whether the statement is True or False. A reciprocal of a fraction is obtained by inverting it upside down.

Answer»

A reciprocal of a fraction is obtained by inverting it upside down.

True

Discussion

21.	Fill in the blanks to make the statements true. 4.7 ÷ 10 =_________
Answer» 4.7 ÷ 10= 0.47

Discussion

22.	State whether the statement is True or False. To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.
Answer» To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places. True

22.

State whether the statement is True or False. To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.

Answer»

To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.

True

Discussion

23.	Fill in the blanks to make the statements true. 3.2 x 10 =_________
Answer» 3.2 x 10 = 32

Discussion

24.	Fill in the blanks to make the statements true. 25.4 x 1000 =__________
Answer» 25.4 x 1000 =25400

Discussion

25.	Curved surface area of a cylinder is(a) πr2h(b) 2πr(h + r)(c) 2πrh(d) 2πr
Answer» Curved surface area of a cylinder is 2πrh.

Discussion

26.	If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2 .
Answer» We know, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) 9² = a² + b² + c² + 2(40) 81 = a² + b² + c² + 80 ⇒ a² + b² + c² = 1

26.

If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2 .

Answer»

We know, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

9² = a² + b² + c² + 2(40)

81 = a² + b² + c² + 80

⇒ a² + b² + c² = 1

Discussion

27.	Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 5m – 3 ii. a iii. 4 iv. 3y² – 7y + 5
Answer» i. Binomial ii. Monomial iii. Monomial iv. Trinomial

27.

Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 5m – 3 ii. a iii. 4 iv. 3y² – 7y + 5

Answer»

i. Binomial

ii. Monomial

iii. Monomial

iv. Trinomial

Discussion

28.	Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 7x ii. 5y – 7z iii. 3x³ – 5x² – 11 iv. 1 – 8a – 7a² – 7a³
Answer» i. Monomial ii. Binomial iii. Trinomial iv. Polynomial

28.

Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 7x ii. 5y – 7z iii. 3x³ – 5x² – 11 iv. 1 – 8a – 7a² – 7a³

Answer»

i. Monomial

ii. Binomial

iii. Trinomial

iv. Polynomial

Discussion

29.	Why did the Bear want to kill a cow?
Answer» The Bear wanted to kill a cow so that the Dog and he could eat to their fill.

Discussion

30.	Why did the Dog want his master to be the strongest?
Answer» The Dog wanted his master to be the strongest as he did not want to be frightened of animals stronger than he.

Discussion

31.	What were the cows doing?
Answer» The cows were mooing loudly and running in panic in sill directions.

Discussion

32.	Who did he finally choose as his master and why?
Answer» The Dog finally chose man as his master as he realized that man is the most powerful being on earth. Even the strongest animal Like the lion was also afraid of him. Thus, his search for the strongest being on the earth was over.

Discussion

33.	Who was the second master of the Dog?
Answer» A Bear was the second master on the Dog.

Discussion

34.	Why did he serve the Lion for a long time?
Answer» The Dog served the Lion for a long time as he was happy and he had nothing to complain of. There was no stronger beast than the Lion in the forest and no one dared to touch the Dog or trouble him In any way.

Discussion

35.	Who did the Dog first choose as his master ? Why did he leave him ?
Answer» The Dog first chose his kinsman-a Wolf-as his master. Once while walking in forest the Wolf smelt a Bear coming and being frightened he left the path and into the bushes and crept deeper into the forest.

Discussion

36.	The Lion felt that he would be in trouble. Why?
Answer» The Lion smelt a man coming and said that he would be in trouble.

Discussion

37.	Who, according to the Bear, was the strongest beast on earth?
Answer» According to the Bear, the Lion was the strongest beast on earth.

Discussion

38.	What kind of master was the Dog looking for?
Answer» The Dog was looking for such a master who was stronger than anyone on the earth.

Discussion

39.	What solution did the Dog think to get rid of his unpleasant life?
Answer» The Dog thought to become the servant of one who was stronger them anyone on the earth.

Discussion

40.	Why did the Dog serve the Lion for a long time?
Answer» The Bear had told the Dog that the Lion was the strongest beast on earth. So the Dog felt good in the company of the Lion and had nothing to complain of. Thus feeling greatly safe and secured, he served the Lion for a long time.

Discussion

41.	Why did the Bear run away when he saw a herd of cows?
Answer» The Bear wanted to kill a cow and eat to his fill. But suddenly the cows started mooing loudly and running in a panic in all directions. They had smelt the Lion coming.

Discussion

42.	What kind of life was the Dog ill-pleased with?
Answer» Once Dogs were their own masters and lived freely as the wolves do. The Dog was ill-pleased with this kind of life.

Discussion

43.	Which two qualities of the Wolf appealed the Dog?
Answer» The Wolf is Dog’s kinsman. The Dog found that the Wolf is strong and fierce as well.

Discussion

44.	The cows were running in a panic for …A. they saw the Dog.B. they saw the Bear.C. they heard the roar of a Lion.D. they heard the snort of a Bear.
Answer» Correct option is C. they heard the roar of a Lion.

Discussion

45.	‘Eat our fill’ means …A. ‘Eat delicious food’.B. ‘Eat when hungry’.C. ‘Eat to the fullest of our capacity’.D. ‘Eat without any interruption’.
Answer» Correct option is C. ‘Eat to the fullest of our capacity’.

Discussion

46.	The Dog wanted to be a ……………… of one who was stronger than anyone on earth.A. masterB. followerC. servantD. friend
Answer» Correct option is C. servant

Discussion

47.	How did the Lion react when he smelt a man approaching him?
Answer» When the Lion smelt a man approaching, he was greatly frightened. He told the Dog that it would be better if they ran away or they would be in trouble.

Discussion

48.	Read the following passages and answer the questions given below them:Dogs were once their own masters and lived the way wolves do, in freedom, until a dog was born who was ill-pleased with his way of life. He was sick and tired of wandering about by himself looking for food and being frightened of those who were stronger than he. He thought it over and decided that the best thing for him to do was to become the servant of one who was stronger than anyone on earth and he set out to find such a master.1. How did the Dogs live in the beginning?2. Who changed the way the Dogs lived?3. What was the Dog tired of?4. What was the Dog frightened of?
Answer» 1. In the beginning, the Dogs were their own masters and lived in freedom like the Wolves do. 2. A Dog who was ill-pleased with his way of life changed the way the Dogs lived.’ 3. The Dog was tired of wandering about s by himself looking for food. 4. The Dog was frightened of those who were stronger than he.

48.

Read the following passages and answer the questions given below them:Dogs were once their own masters and lived the way wolves do, in freedom, until a dog was born who was ill-pleased with his way of life. He was sick and tired of wandering about by himself looking for food and being frightened of those who were stronger than he. He thought it over and decided that the best thing for him to do was to become the servant of one who was stronger than anyone on earth and he set out to find such a master.1. How did the Dogs live in the beginning?2. Who changed the way the Dogs lived?3. What was the Dog tired of?4. What was the Dog frightened of?

Answer»

1. In the beginning, the Dogs were their own masters and lived in freedom like the Wolves do.

2. A Dog who was ill-pleased with his way of life changed the way the Dogs lived.’

3. The Dog was tired of wandering about s by himself looking for food.

4. The Dog was frightened of those who were stronger than he.

Discussion

49.	Read the following passages and answer the questions given below them:But one day the two of -them were walking side by side along a path that ran amid bare cliffs when all of a sudden the Lion stopped. He gave a great roar and struck the ground angrily with his paw with such force that a hole formed there. Then ? he began to back away very quietly. “What is it, Master, is anything wrong?” asked the Dog, surprised.“I smell a man coming this way,” the Lion said, “We’d better run for it or we’ll be in trouble.” “Oh, well, then I’ll say goodbye to you, Lion. I want a master who is stronger than anyone on earth!”And off the Dog went to join the man and stayed with him and served him faithfully. This happened long, long ago, but to this day the Dog is man’s most loyal servant and knows no other master.1. What did the Lion do after stopping?2. Why did the Lion back away quietly?3. Why did the Dog say goodbye to the Lion?4. Why does the Dog remain man’s most faithful servant?
Answer» 1. Once the Lion stopped, he gave out a great roar and struck the ground angrily with his paw with such force forming a hole there. 2. The Lion backed away quietly because he smelled a man and thought that he would be in trouble. 3. The Dog realized that the man was : stronger than the Lion. Since the Dog wanted s the strongest master, he said goodbye to the ) Lion and went to the man. 4. The Dog remains man’s most faithful servant as the Dog has not found anyone stronger than man on earth.

49.

Read the following passages and answer the questions given below them:But one day the two of -them were walking side by side along a path that ran amid bare cliffs when all of a sudden the Lion stopped. He gave a great roar and struck the ground angrily with his paw with such force that a hole formed there. Then ? he began to back away very quietly. “What is it, Master, is anything wrong?” asked the Dog, surprised.“I smell a man coming this way,” the Lion said, “We’d better run for it or we’ll be in trouble.” “Oh, well, then I’ll say goodbye to you, Lion. I want a master who is stronger than anyone on earth!”And off the Dog went to join the man and stayed with him and served him faithfully. This happened long, long ago, but to this day the Dog is man’s most loyal servant and knows no other master.1. What did the Lion do after stopping?2. Why did the Lion back away quietly?3. Why did the Dog say goodbye to the Lion?4. Why does the Dog remain man’s most faithful servant?

Answer»

1. Once the Lion stopped, he gave out a great roar and struck the ground angrily with his paw with such force forming a hole there.

2. The Lion backed away quietly because he smelled a man and thought that he would be in trouble.

3. The Dog realized that the man was : stronger than the Lion. Since the Dog wanted s the strongest master, he said goodbye to the ) Lion and went to the man.

4. The Dog remains man’s most faithful servant as the Dog has not found anyone stronger than man on earth.

Discussion

50.	The Wolf is Dog’s kinsman as …A. it looks like a Dog.B. it belongs to the same species as the Dog.C. it is omnivorous as the Dog.D. None of the above
Answer» Correct option is B. it belongs to the same species as the Dog.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

What does the law do to unjust customs?

There are several ………………….. customs and traditions in our society. (a) good (b) bad (c) ugly

What are the differences between “Logical error” and “Syntax error”?

In Fig. ABCD is a trapezium in which AB||DC. Prove that ar(Δ AOD) = ar(Δ BOC).

In Fig. X and Y are the mid-point of AC and AB respectively, QP||BC and CYQ and BXP are straight lines. Prove that : ar(Δ ABP) = ar(Δ ACQ).

In a Δ ABC, if L and M are points on AB and AC respectively such that LM||BC. Prove that : (i) ar(ΔLCM) = ar(ΔLBM) (ii) ar(ΔLBC) = ar(ΔMBC) (iii) ar(ΔABM) = ar(ΔACL) (iv) ar(ΔLOB) = ar(ΔMOC)

In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP. Prove that : (i) ar(ΔPBQ) = ar(ΔARC) (ii) ar(ΔPQR) = \(\frac{1}{2}\)ar(ΔARC) (iii) ar(ΔRQC) = \(\frac{3}{8}\)ar(ΔABC)

D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, Prove that, ar(ΔBOE) = \(\frac{1}{8}\)ar(ΔABC)

In Fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Compute the area of trapezium PQRS in Fig.

In Fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

In figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.

Simplify and write the result in decimal form :(1÷ 2/9) + (1 ÷ 3 1/5) + (1 ÷ 2 2/3)

Fill in the blanks to make the statement true.To divide a decimal number by 100, we shift the decimal point in the number to the ________ by ______ places.

Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

To find the distance around a circular disc, multiply the diameter of the disc by 3.14. What is the distance around the disc when :(a) the diameter is 18.7 cm?(b) the radius is 6.45 cm?

What is the Error in the question. A student compared – 1/4 and –0.3. He changed –1/ 4 to the decimal –0.25 and wrote, “Since 0.3 is greater than 0.25, –0.3 is greater than –0.25”. What was the student’s error?

A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by 1.414. Find the length of the diagonal when :(a) The length of a side of the square is 8.3 cm.(b) The length of a side of the square is exactly 7.875 cm.

State whether the statement is True or False. A reciprocal of a fraction is obtained by inverting it upside down.

Fill in the blanks to make the statements true. 4.7 ÷ 10 =_________

State whether the statement is True or False. To multiply a decimal number by 1000, we move the decimal point in the number to the right by three places.

Fill in the blanks to make the statements true. 3.2 x 10 =_________

Fill in the blanks to make the statements true. 25.4 x 1000 =__________

Curved surface area of a cylinder is(a) πr2h(b) 2πr(h + r)(c) 2πrh(d) 2πr

If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2 .

Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 5m – 3 ii. a iii. 4 iv. 3y² – 7y + 5

Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.i. 7x ii. 5y – 7z iii. 3x³ – 5x² – 11 iv. 1 – 8a – 7a² – 7a³

Why did the Bear want to kill a cow?

Why did the Dog want his master to be the strongest?

What were the cows doing?

Who did he finally choose as his master and why?

Who was the second master of the Dog?

Why did he serve the Lion for a long time?

Who did the Dog first choose as his master ? Why did he leave him ?

The Lion felt that he would be in trouble. Why?

Who, according to the Bear, was the strongest beast on earth?

What kind of master was the Dog looking for?

What solution did the Dog think to get rid of his unpleasant life?

Why did the Dog serve the Lion for a long time?

Why did the Bear run away when he saw a herd of cows?

What kind of life was the Dog ill-pleased with?

Which two qualities of the Wolf appealed the Dog?

The cows were running in a panic for …A. they saw the Dog.B. they saw the Bear.C. they heard the roar of a Lion.D. they heard the snort of a Bear.

‘Eat our fill’ means …A. ‘Eat delicious food’.B. ‘Eat when hungry’.C. ‘Eat to the fullest of our capacity’.D. ‘Eat without any interruption’.

The Dog wanted to be a ……………… of one who was stronger than anyone on earth.A. masterB. followerC. servantD. friend

How did the Lion react when he smelt a man approaching him?

The Wolf is Dog’s kinsman as …A. it looks like a Dog.B. it belongs to the same species as the Dog.C. it is omnivorous as the Dog.D. None of the above

Fill in the blanks to make the statement true.To divide a decimal number by 100, we shift the decimal point in the number to the __ by places.