125876 + Interview Questions in General QA Page 517 InterviewSolution

25801.	Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.(A) 16(B) 56(C) 24(D) 8
Answer» (B) 56 A triangle is obtained by joining three vertices. Number of ways of selecting 3 vertices out of 8 vertices = ⁸C₃ \(=\frac{8\times7\times6}{1\times2\times3}\) = 56

25801.

Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.(A) 16(B) 56(C) 24(D) 8

Answer»

(B) 56

A triangle is obtained by joining three vertices. Number of ways of selecting 3 vertices out of 8 vertices = ⁸C₃

\(=\frac{8\times7\times6}{1\times2\times3}\) = 56

Discussion

25802.	Find the value of\(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\)
Answer» Given expression is \(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\) \(=\space^{56}C_4+^{55}C_3+^{54}C_3+^{53}C_3+^{52}C_3+^{51}C_3+^{50}C_3\) Writing the terms in reverse order, we get =\((^{50}C_4 +^{50}C_3)+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)= \(^{50}C_4+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\) [∴ \(^nC_r+^nC_{r-1}=^{n+1}C_r\)] = \((^{51}C_4+^{51}C_3)+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\) = \((^{52}C_4+^{52}C_3)+^{53}C_3+^{54}C_3+^{55}C_3\) = \((^{53}C_4+^{53}C_3)+^{54}C_3+^{55}C_3\) = \((^{54}C_4+^{54}C_3)+^{55}C_3\) = \(^{55}C_4+^{55}C_3\) = \(^{56}C_4\)

25802.

Find the value of\(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\)

Answer»

Given expression is

\(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\)

\(=\space^{56}C_4+^{55}C_3+^{54}C_3+^{53}C_3+^{52}C_3+^{51}C_3+^{50}C_3\)

Writing the terms in reverse order, we get

=\((^{50}C_4 +^{50}C_3)+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)= \(^{50}C_4+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)

[∴ \(^nC_r+^nC_{r-1}=^{n+1}C_r\)]

= \((^{51}C_4+^{51}C_3)+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)

= \((^{52}C_4+^{52}C_3)+^{53}C_3+^{54}C_3+^{55}C_3\)

= \((^{53}C_4+^{53}C_3)+^{54}C_3+^{55}C_3\)

= \((^{54}C_4+^{54}C_3)+^{55}C_3\)

= \(^{55}C_4+^{55}C_3\)

= \(^{56}C_4\)

Discussion

25803.	The set S = {1,2,3,…,12} is to be partitioned into three sets A,B, and C of equal sizes. A ⋃ B ⋃ C = S, A ⋂ B = B ⋂ C = C ⋂ A = Φ Find the number of ways to partitions.
Answer» There are 12 elements. Three disjoint sets A,B and C of same size are to be formed. ⇒ Each set has 4 elements. Number of ways of selecting any 4 elements from 12 elements for set A = ¹²C₄ Number of ways of selecting any 4 elements from 8 elements for set B = ⁸C₄ Number of ways of selecting any 4 elements from 4 elements for C = ⁴C₄ ∴ Total number of partitions = ¹²C₄ × ⁸C₄ × ⁴C₄

25803.

The set S = {1,2,3,…,12} is to be partitioned into three sets A,B, and C of equal sizes. A ⋃ B ⋃ C = S, A ⋂ B = B ⋂ C = C ⋂ A = Φ Find the number of ways to partitions.

Answer»

There are 12 elements.

Three disjoint sets A,B and C of same size are to be formed.

⇒ Each set has 4 elements.

Number of ways of selecting any 4 elements from 12 elements for set A = ¹²C₄

Number of ways of selecting any 4 elements from 8 elements for set B = ⁸C₄

Number of ways of selecting any 4 elements from 4 elements for C = ⁴C₄

∴ Total number of partitions = ¹²C₄ × ⁸C₄ × ⁴C₄

Discussion

25804.	Show that the Lagrange's mean value theorem is not applicable to the function f(x) = 1/x on [–1, 1].
Answer» Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that f(b) − f(a) = f′(c)(b − a) \(\Rightarrow f'(c) = \frac{f(b) - f(a)}{b - a}\) This theorem is also known as First Mean Value Theorem. f(x) = \(\frac{1}{x}\) on [-1, 1] Here, x ≠ 0 ⇒ f(x) exists for all values of x except 0 ⇒ f(x) is discontinuous at x = 0 ∴ f(x) is not continuous in [– 1, 1] Hence the lagrange’s mean value theorem is not applicable to the function f(x) = \(\frac{1}{x}\) on [-1, 1]

25804.

Show that the Lagrange's mean value theorem is not applicable to the function f(x) = 1/x on [–1, 1].

Answer»

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that

f(b) − f(a) = f′(c)(b − a)

\(\Rightarrow f'(c) = \frac{f(b) - f(a)}{b - a}\)

This theorem is also known as First Mean Value Theorem.

f(x) = \(\frac{1}{x}\) on [-1, 1]

Here,

x ≠ 0

⇒ f(x) exists for all values of x except 0

⇒ f(x) is discontinuous at x = 0

∴ f(x) is not continuous in [– 1, 1]

Hence the lagrange’s mean value theorem is not applicable to the

function f(x) = \(\frac{1}{x}\) on [-1, 1]

Discussion

25805.	In how many ways can 5 children be arranged in a line such that (i) two particular children of them are always together (ii) two particular children of them are never together.
Answer» (i) We consider the arrangements by taking 2 particular children together as one and hence the remaining 4 can be arranged in 4! = 24 ways. Again two particular children taken together can be arranged in two ways. Therefore, there are 24 × 2 = 48 total ways of arrangement. (ii) Among the 5! = 120 permutations of 5 children, there are 48 in which two children are together. In the remaining 120 – 48 = 72 permutations, two particular children are never together.

25805.

In how many ways can 5 children be arranged in a line such that (i) two particular children of them are always together (ii) two particular children of them are never together.

Answer»

(i) We consider the arrangements by taking 2 particular children together as one and hence the remaining 4 can be arranged in 4! = 24 ways. Again two particular children taken together can be arranged in two ways. Therefore, there are 24 × 2 = 48 total ways of arrangement.

(ii) Among the 5! = 120 permutations of 5 children, there are 48 in which two children are together. In the remaining 120 – 48 = 72 permutations, two particular children are never together.

Discussion

25806.	The number of possible outcomes when a coin is tossed 6 times is(A) 36 (B) 64 (C) 12(D) 32
Answer» Answer is (B) Number of outcomes when a coin tossed = 2 (Head or Tail) ∴Total possible outcomes when a coin tossed 6 times = 2 x 2 x 2 x 2 x 2 x 2 = 64

25806.

The number of possible outcomes when a coin is tossed 6 times is(A) 36 (B) 64 (C) 12(D) 32

Answer»

Answer is (B)

Number of outcomes when a coin tossed = 2 (Head or Tail)

∴Total possible outcomes when a coin tossed 6 times = 2 x 2 x 2 x 2 x 2 x 2 = 64

Discussion

25807.	If nC8 = nC12 find nC5.
Answer» Given ⁿC₈ = ⁿC₁₂ ⇒ n = 8 + 12 = 20 ⁿC₅ = ²⁰C₅ = 15,504.

25807.

If nC8 = nC12 find nC5.

Answer»

Given ⁿC₈ = ⁿC₁₂

⇒ n = 8 + 12 = 20

ⁿC₅ = ²⁰C₅ = 15,504.

Discussion

25808.	In how many ways can a football team of 11 players be selected from 16 players?How many of them will(i) include 2 particular players?(ii) exclude 2 particular players?
Answer» Total number of players = 16 We have to select a team of 11 players So, number of ways = ¹⁶C₁₁ (i) If two particular players are included then more 9 players can be selected from remaining 14 players in ¹⁴C₉ (ii) If two particular players are excluded then all 11 players can be selected from remaining 14 players in ¹⁴C₁₁

25808.

In how many ways can a football team of 11 players be selected from 16 players?How many of them will(i) include 2 particular players?(ii) exclude 2 particular players?

Answer»

Total number of players = 16

We have to select a team of 11 players

So, number of ways = ¹⁶C₁₁

(i) If two particular players are included then more 9 players can be selected from remaining 14 players in ¹⁴C₉

(ii) If two particular players are excluded then all 11 players can be selected from remaining 14 players in ¹⁴C₁₁

Discussion

25809.	In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only?(a) 112 (b) 132 (c) 91 (d) 182
Answer» (b) 132 As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. So the number of ways in which a team of 11 can be selected = ¹²C₁₀ x ²C₁ \(\frac{12!}{10!\times2!}\times2\) = \(\frac{12\times11}{2}\times2\) = 132.

25809.

In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only?(a) 112 (b) 132 (c) 91 (d) 182

Answer»

(b) 132

As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members.

So the number of ways in which a team of 11 can be selected

= ¹²C₁₀ x ²C₁ \(\frac{12!}{10!\times2!}\times2\)

= \(\frac{12\times11}{2}\times2\) = 132.

Discussion

25810.	A team of 8 players has to be selected from 14 players. In how many ways the selections can be made if (a) Two particular players are always included. (b) Two particular players are always excluded.
Answer» (a) Two players are always included: We have to choose 6 from 12 players. This can be done in ¹²C₆ ways = 924 ways. (b) Two players are always excluded: We have to choose 8 from 12 players. This can be done in ¹²C₈ ways = 495 ways.

25810.

A team of 8 players has to be selected from 14 players. In how many ways the selections can be made if (a) Two particular players are always included. (b) Two particular players are always excluded.

Answer»

(a) Two players are always included:

We have to choose 6 from 12 players. This can be done in ¹²C₆ ways = 924 ways.

(b) Two players are always excluded:

We have to choose 8 from 12 players. This can be done in ¹²C₈ ways = 495 ways.

Discussion

25811.	If npr = 240, nCr = 120 find n and r.
Answer» Given ⁿp_r = 240, ⁿC_r = 120 w.k.t ncr = \(\frac{^nP_r}{r!}\) ∴ r! = \(\frac{240}{120}\) = 2 r! = 2! ⇒ r= 2.

25811.

If npr = 240, nCr = 120 find n and r.

Answer»

Given ⁿp_r = 240, ⁿC_r = 120

w.k.t ncr = \(\frac{^nP_r}{r!}\)

∴ r! = \(\frac{240}{120}\) = 2

r! = 2! ⇒ r= 2.

Discussion

25812.	A letter lock contains 3 rings and each ring contains 5 letters. Determine the maximum number of false trails that can be made before the lock is opened.
Answer» A letter lock has 3 rings, each ring containing 5 different letters. ∴ A letter from each ring can be selected in 5 ways. ∴ By using the fundamental principle of multiplication, a total number of trials that can be made = 5 × 5 × 5 = 125. Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened. ∴ A maximum number of false trials = 124

25812.

A letter lock contains 3 rings and each ring contains 5 letters. Determine the maximum number of false trails that can be made before the lock is opened.

Answer»

A letter lock has 3 rings, each ring containing 5 different letters.

∴ A letter from each ring can be selected in 5 ways.

∴ By using the fundamental principle of multiplication,

a total number of trials that can be made = 5 × 5 × 5 = 125.

Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened.

∴ A maximum number of false trials = 124

Discussion

25813.	A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer» Given, Height of the cylindrical bucket = 32 cm Radius of the cylindrical bucket = 18 cm Height of conical heap = 24 cm We know that, Volume of cylinder = π × r² × h And, volume of cone = 1/3 π × r² × h Then, from the question Volume of the conical heap = Volume of the cylindrical bucket 1/3 π × r² × 24 = π × 18² × 32 r² = 18² x 4 r = 18 x 2 = 36 cm Now, Slant height of the conical heap (l) is given by l = √(h² + r²) l = √(24² + 36²) = √1872 l = 43.26 cm Therefore, the radius and slant height of the conical heap are 36 cm and 43.26 cm respectively.

25813.

A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer»

Given,

Height of the cylindrical bucket = 32 cm

Radius of the cylindrical bucket = 18 cm

Height of conical heap = 24 cm

We know that,

Volume of cylinder = π × r² × h

And, volume of cone = 1/3 π × r² × h

Then, from the question

Volume of the conical heap = Volume of the cylindrical bucket

1/3 π × r² × 24 = π × 18² × 32

r² = 18² x 4

r = 18 x 2 = 36 cm

Now,

Slant height of the conical heap (l) is given by

l = √(h² + r²)

l = √(24² + 36²) = √1872

l = 43.26 cm

Therefore, the radius and slant height of the conical heap are 36 cm and 43.26 cm respectively.

Discussion

25814.	A floor is 4.5 m long and 3.6 m wide. A 6 cm square tile costs Rs 23.25. What will be the cost to cover the floor with these tiles?
Answer» Correct answer is 104625

Discussion

25815.	Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Answer» Given that, Let the length of side of trapezium shaped field along road = x meter Length of other side of trapezium shaped field along road = 2x meter Area of trapezium = 10500 cm² Distance between parallel sides = 100 m We know that, Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel sides i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides 10500 = 1/2 (x + 2x) × 100 10500 = 1/2 (3x) × 100 3x = 10500/50 3x = 210 x = 210/3 = 70 x = 70 ∴ Length of side of trapezium shaped field along road = 70 m And, Length of other side of trapezium shaped field along road = 2x = 70× 2 = 140 m

25815.

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer»

Given that,

Let the length of side of trapezium shaped field along road = x meter

Length of other side of trapezium shaped field along road = 2x meter

Area of trapezium = 10500 cm²

Distance between parallel sides = 100 m

We know that,

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel sides

i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides

10500 = 1/2 (x + 2x) × 100

10500 = 1/2 (3x) × 100

3x = 10500/50

3x = 210

x = 210/3 = 70

x = 70

∴ Length of side of trapezium shaped field along road = 70 m

And, Length of other side of trapezium shaped field along road = 2x = 70× 2 = 140 m

Discussion

25816.	The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
Answer» Given Let length of first parallel side X Length of other parallel side is X + 6 Area of trapezium= 180 cm² Let Height (h) = 9 cm We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height Therefore Area of trapezium is \(\frac{1}{2}\)× (X + 6 +X) × 9 = 180 cm². \(\therefore\) \(\frac{1}{2}\)× (X + 6 +X) × 9 =180 ⇒ \(\frac{1}{2}\)× (2X + 6) × 9 =180 ⇒ 2X + 6 = \(\frac{180}{9}\times2\) ⇒ 2X + 6 = 40 ⇒ 2X = 40 – 6 = 34 ⇒ X = 17 ∴ Length of the parallel sides is X = 17 cm and X + 6 = 17 + 6 = 23 cm. Therefore lengths of the parallel sides are 17 cm, 23 cm.

25816.

The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.

Answer»

Given

Let length of first parallel side X

Length of other parallel side is X + 6

Area of trapezium= 180 cm²

Let Height (h) = 9 cm

We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height

Therefore Area of trapezium is \(\frac{1}{2}\)× (X + 6 +X) × 9 = 180 cm².

\(\therefore\) \(\frac{1}{2}\)× (X + 6 +X) × 9 =180

⇒ \(\frac{1}{2}\)× (2X + 6) × 9 =180

⇒ 2X + 6 = \(\frac{180}{9}\times2\)

⇒ 2X + 6 = 40

⇒ 2X = 40 – 6 = 34

⇒ X = 17

∴ Length of the parallel sides is X = 17 cm and X + 6 = 17 + 6 = 23 cm.

Therefore lengths of the parallel sides are 17 cm, 23 cm.

Discussion

25817.	The parallel sides of a trapezium measure 14cm and 18cm and the distance between them is 9 cm. The area of the trapezium is
Answer» We know that the given details are: Length of the parallel sides are 14cm and 18cm Height (h) = 9cm By using the formula, Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (14+18) × 9 =144cm²

25817.

The parallel sides of a trapezium measure 14cm and 18cm and the distance between them is 9 cm. The area of the trapezium is

Answer»

We know that the given details are:

Length of the parallel sides are 14cm and 18cm

Height (h) = 9cm

By using the formula,

Area of trapezium = ½ × (sum of parallel sides) × height

= ½ × (14+18) × 9

=144cm²

Discussion

25818.	Find the area of hexagon, ABCDEF in which BL ⊥ AD, CM ⊥ AD and EN ⊥ AD and FP ⊥ AD such that AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm.
Answer» The given details are, BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm. AL=AP+PL =6+2 = 8cm PN= PL+LN= 2+8 =10cm LM=LN+NM=8+2=10cm ND=NM+MD=2+3=5cm By using the formula, Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB) Area of triangle= ½ × base × height Area of trapezium = ½ × (sum of parallel sides) × height ∴ lets calculate, Area (△APF) = ½ (AP) × (FP) = ½ × 6 × 8 = 24cm² Area (△DEN) =½ (ND) × (EN) = ½ × 5 × 12 = 30cm² Area (△ABL) =½ (AL) × (BL) = ½ × 8 × 8 = 32cm² Area (△CMD) =½ (MD) × (CM) = ½ × 3 × 6 = 9cm² Area (Trap.PNEF) =½ × (FP + EN) × PN = ½ × (18+12) × 10 = 100cm² Area (Trap.LMCB) =½ × (BL + CM) × LM = ½ × (8+6) × 10 = 70cm² ∴ Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB) = 24 + 30 + 32 + 9+100+70 = 265cm²

25818.

Find the area of hexagon, ABCDEF in which BL ⊥ AD, CM ⊥ AD and EN ⊥ AD and FP ⊥ AD such that AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm.

Answer»

The given details are,

BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD

AP=6cm, PL=2cm, LN=8cm, NM=2cm, MD=3cm, FP=8cm, EN=12cm, BL=8cm, CM=6cm.

AL=AP+PL =6+2 = 8cm

PN= PL+LN= 2+8 =10cm

LM=LN+NM=8+2=10cm

ND=NM+MD=2+3=5cm

By using the formula,

Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB)

Area of triangle= ½ × base × height

Area of trapezium = ½ × (sum of parallel sides) × height

∴ lets calculate,

Area (△APF) = ½ (AP) × (FP) = ½ × 6 × 8 = 24cm²

Area (△DEN) =½ (ND) × (EN) = ½ × 5 × 12 = 30cm²

Area (△ABL) =½ (AL) × (BL) = ½ × 8 × 8 = 32cm²

Area (△CMD) =½ (MD) × (CM) = ½ × 3 × 6 = 9cm²

Area (Trap.PNEF) =½ × (FP + EN) × PN = ½ × (18+12) × 10 = 100cm²

Area (Trap.LMCB) =½ × (BL + CM) × LM = ½ × (8+6) × 10 = 70cm²

∴ Area (hex.ABCDEF) = area (△APF) + area (△DEN) + area (△ABL) + area (△CMD) + area (Trap.PNEF) + area (Trap.LMCB)

= 24 + 30 + 32 + 9+100+70

= 265cm²

Discussion

25819.	In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m2 and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.
Answer» Given Let length of first parallel side X Length of other parallel side is 2X Area of trapezium= 9450 m² Let Height (h) = 84 m We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height Therefore Area of trapezium is \(\frac{1}{2}\)× (X + 2X) × 84 = 9450cm². \(\therefore\) \(\frac{1}{2}\)× (X + 2X) × 84 =9450 ⇒ (3X) × 42 =9450 ⇒ 126X = 9450 ⇒ 2X + 6 = \(\frac{9450}{126}=75\) ⇒ X = 17 ∴ Length of the parallel sides is X = 75 m and 2X = 150 m. Therefore length of the longest is 150 m.

25819.

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m2 and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.

Answer»

Given

Let length of first parallel side X

Length of other parallel side is 2X

Area of trapezium= 9450 m²

Let Height (h) = 84 m

We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height

Therefore Area of trapezium is \(\frac{1}{2}\)× (X + 2X) × 84 = 9450cm².

\(\therefore\) \(\frac{1}{2}\)× (X + 2X) × 84 =9450

⇒ (3X) × 42 =9450

⇒ 126X = 9450

⇒ 2X + 6 = \(\frac{9450}{126}=75\)

⇒ X = 17

∴ Length of the parallel sides is X = 75 m and 2X = 150 m.

Therefore length of the longest is 150 m.

Discussion

25820.	The area of a trapezium is 405 cm2. Its parallel sides are in the ration 4:5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
Answer» we know that the length of the parallel sides is 4xcm and 5xcm Height (h) = 18cm Area of trapezium = 405cm² By using the formula Area of trapezium = ½ × (sum of parallel sides) × height 405 = ½ × (4x+5x) × 18 405 = ½ × (4x+5x) × 18 405 = (9x) × 9 9x = 405/9 x= 45/9 x = 5 ∴ The length of the other parallel side is 4x= 4 × 5 = 20 cm and 5x= 5 × 5 = 25 cm

25820.

The area of a trapezium is 405 cm2. Its parallel sides are in the ration 4:5 and the distance between them is 18 cm. Find the length of each of the parallel sides.

Answer»

we know that the length of the parallel sides is 4xcm and 5xcm

Height (h) = 18cm

Area of trapezium = 405cm²

By using the formula Area of trapezium = ½ × (sum of parallel sides) × height

405 = ½ × (4x+5x) × 18

405 = (9x) × 9

9x = 405/9

x= 45/9

x = 5

∴ The length of the other parallel side is 4x= 4 × 5 = 20 cm and 5x= 5 × 5 = 25 cm

Discussion

25821.	The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54m, CD = 19m and AD = 42m, find the area of the field.
Answer» Given Length of parallel sides AD = 42 m BC = 54 m Given that total length of fence is 130 m That is AB + BC +CD + DA = 130 AB + 54 + 19 + 42 = 130 Therefore AB = 15 Height (AB) = 15 m We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height Therefore Area of trapezium = \(\frac{1}{2}\)× (42 + 54) × 15 = 720 m²

25821.

The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54m, CD = 19m and AD = 42m, find the area of the field.

Answer»

Given

Length of parallel sides

AD = 42 m

BC = 54 m

Given that total length of fence is 130 m

That is AB + BC +CD + DA = 130

AB + 54 + 19 + 42 = 130

Therefore AB = 15

Height (AB) = 15 m

We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height

Therefore Area of trapezium = \(\frac{1}{2}\)× (42 + 54) × 15 = 720 m²

Discussion

25822.	In the given figure, ABCD is a quadrilateral in which AC=24cm, BL ⊥ AC and DM ⊥ AC such that BL=8cm and DM=7cm. find the area of quad. ABCD.
Answer» The given details are, BL ⊥ AC and DM ⊥ AC AC=24cm, BL=8cm, DM=7cm By using the formula, Area (quad.ABCD) = area (△ABC) + area(△ADC) Area of triangle=1/2 × base × height ∴ Area (quad.ABCD) = (½ × AC × BL) + (½ × AC × DM) = (½ × 24 × 8) + (½ × 24 × 7) = 96 + 84 = 180cm²

25822.

In the given figure, ABCD is a quadrilateral in which AC=24cm, BL ⊥ AC and DM ⊥ AC such that BL=8cm and DM=7cm. find the area of quad. ABCD.

Answer»

The given details are,

BL ⊥ AC and DM ⊥ AC

AC=24cm, BL=8cm, DM=7cm

By using the formula,

Area (quad.ABCD) = area (△ABC) + area(△ADC)

Area of triangle=1/2 × base × height

∴ Area (quad.ABCD) = (½ × AC × BL) + (½ × AC × DM)

= (½ × 24 × 8) + (½ × 24 × 7)

= 96 + 84

= 180cm²

Discussion

25823.	Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.
Answer» Given: A pentagon ABCDE BL \(\perp\) AC, CM \(\perp\) AD and EN \(\perp\) AD AC = 10 cm AD = 12 cm BL = 3 cm CM = 7 cm EN = 5 cm Here, Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) Area of triangle = \(\frac{1}{2}\)× (base) × (height). Here, Area (\(\triangle\)ABC) = \(\frac{1}{2}\)× (AC) × (BL) = \(\frac{1}{2}\)× (10) × (3) = 15 cm². Area (\(\triangle\)ACD) = \(\frac{1}{2}\)× (AD) × (CD) = \(\frac{1}{2}\)× (12) × (7) = 42 cm². Area (\(\triangle\)ADE) = \(\frac{1}{2}\)× (AD) × (EN) = \(\frac{1}{2}\)× (12) × (5) = 30 cm². ∴ Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) = 15 + 42 + 30 = 87 cm². ∴ Area (Pent. ABCDE) = 87 cm².

25823.

Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

Answer»

Given: A pentagon ABCDE

BL \(\perp\) AC, CM \(\perp\) AD and EN \(\perp\) AD

AC = 10 cm

AD = 12 cm

BL = 3 cm

CM = 7 cm

EN = 5 cm

Here,

Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE)

Area of triangle = \(\frac{1}{2}\)× (base) × (height).

Here,

Area (\(\triangle\)ABC) = \(\frac{1}{2}\)× (AC) × (BL) = \(\frac{1}{2}\)× (10) × (3) = 15 cm².

Area (\(\triangle\)ACD) = \(\frac{1}{2}\)× (AD) × (CD) = \(\frac{1}{2}\)× (12) × (7) = 42 cm².

Area (\(\triangle\)ADE) = \(\frac{1}{2}\)× (AD) × (EN) = \(\frac{1}{2}\)× (12) × (5) = 30 cm².

∴ Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) = 15 + 42 + 30 = 87 cm².

∴ Area (Pent. ABCDE) = 87 cm².

Discussion

25824.	In the given figure, ABCD is a quadrilateral shaped field in which diagonal BD is 36m, AL ⊥ BD and CM ⊥ BD such that AL=19m and CM=11m. Find the area of the field.
Answer» The given details are, AL ⊥ BD and CM ⊥ BD AL=19m, CM=11m, BD=36m By using the formula, Area (quad.ABCD) = area (△ABD) + area(△ACD) Area of triangle=1/2 × base × height ∴ Area (quad.ABCD) = (½ × BD × AL) + (½ × BD × CM) = (½ × 36 × 19) + (½ × 36 × 11) = 342 + 198 = 540m²

25824.

In the given figure, ABCD is a quadrilateral shaped field in which diagonal BD is 36m, AL ⊥ BD and CM ⊥ BD such that AL=19m and CM=11m. Find the area of the field.

Answer»

The given details are,

AL ⊥ BD and CM ⊥ BD

AL=19m, CM=11m, BD=36m

By using the formula,

Area (quad.ABCD) = area (△ABD) + area(△ACD)

Area of triangle=1/2 × base × height

∴ Area (quad.ABCD) = (½ × BD × AL) + (½ × BD × CM)

= (½ × 36 × 19) + (½ × 36 × 11)

= 342 + 198

= 540m²

Discussion

25825.	In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19m and CM = 11m. Find the area of the field.
Answer» Given: A quadrilateral ABCD \(AL\perp BD\) and \(CM\perp BD\) AL = 19 cm BD = 36 cm CM = 11 cm Here, Area (quad. ABCD) = area (△ABD) + area (△ACD) Area of triangle = \(\frac{1}{2}\)× (base) × (height). Therefore Area of quad ABCD = \(\frac{1}{2}\)× (BD) × (AL) + \(\frac{1}{2}\)× (BD) × (CM) = \(\frac{1}{2}\)× (36) × (19) + \(\frac{1}{2}\)× (36) × (11) = 342 + 198 = 540 cm² Therefore area of the quadrilateral ABCD is 540 cm².

25825.

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19m and CM = 11m. Find the area of the field.

Answer»

Given: A quadrilateral ABCD

\(AL\perp BD\) and \(CM\perp BD\)

AL = 19 cm

BD = 36 cm

CM = 11 cm

Here,

Area (quad. ABCD) = area (△ABD) + area (△ACD)

Area of triangle = \(\frac{1}{2}\)× (base) × (height).

Therefore

Area of quad ABCD = \(\frac{1}{2}\)× (BD) × (AL) + \(\frac{1}{2}\)× (BD) × (CM)

= \(\frac{1}{2}\)× (36) × (19) + \(\frac{1}{2}\)× (36) × (11) = 342 + 198 = 540 cm²

Therefore area of the quadrilateral ABCD is 540 cm².

Discussion

25826.	Ramesh’s father is 27 years older than Ramesh. After 5 years, the ratio of ages of Ramesh and his father would be 2 : 3. Find their present age.
Answer» Let age of Ramesh = x years Then father’s age = (x + 27) years After 5 years from now, age of Ramesh = (x + 5) years After 5 years from now, age of father = x+ 27 + 5 = (x+ 32) years According to question, \(\frac{x+5}{x+32} = \frac{2}{3}\) or 3(x + 5) = 2(x + 32) or 3x + 15 = 2x + 64 or 3x – 2x = 64 – 15 or x = 49 Ramesh’s age x = 49 years Father’s age x + 27 = 49 + 27 = 76 years

25826.

Ramesh’s father is 27 years older than Ramesh. After 5 years, the ratio of ages of Ramesh and his father would be 2 : 3. Find their present age.

Answer»

Let age of Ramesh = x years

Then father’s age = (x + 27) years

After 5 years from now, age of Ramesh = (x + 5) years

After 5 years from now, age of father = x+ 27 + 5 = (x+ 32) years

According to question, \(\frac{x+5}{x+32} = \frac{2}{3}\)

or 3(x + 5) = 2(x + 32)

or 3x + 15 = 2x + 64

or 3x – 2x = 64 – 15

or x = 49

Ramesh’s age x = 49 years

Father’s age x + 27 = 49 + 27 = 76 years

Discussion

25827.	Find the three consecutive numbers where the sum of twice of first number, thrice of second number and four times of third number is 182.
Answer» Let three consecutive numbers be x, x + 1 and x + 2. Then according to question 2x + 3(x + 1) + 4 (x + 2) = 182 ⇒ 2x + 3x + 3 + 4x + 8 – 182 ⇒ 9x + 11 = 182 ⇒ 9x = 182 – 11 ⇒ 9x = 171 ⇒ x = 171/9 = 19 Hence, the first number = 19 second number = 19 + 1 = 20 and third number = 19 + 2 = 21

25827.

Find the three consecutive numbers where the sum of twice of first number, thrice of second number and four times of third number is 182.

Answer»

Let three consecutive numbers be x, x + 1 and x + 2.

Then according to question 2x + 3(x + 1) + 4 (x + 2) = 182

⇒ 2x + 3x + 3 + 4x + 8 – 182

⇒ 9x + 11 = 182

⇒ 9x = 182 – 11

⇒ 9x = 171

⇒ x = 171/9 = 19

Hence, the first number = 19

second number = 19 + 1 = 20

and third number = 19 + 2 = 21

Discussion

25828.	Ramesh distributed his capital, half of the Capital to his wife, one third to his son and remaining 50,000/- to his daughter. Find the total amount of his capital.
Answer» Let x be the total money. Then, according to question, Money given to wife = x/2 Money given to son = x/3 Money given to daughter = Rs 50,000 ∴From question, x/2 + x/3 + 50000 = x ⇒ 3x + 2x + 50,000 x 6 = 6x [Multiplying by 6] ⇒ 5x + 3,00,000 = 6x ⇒ 3,00,000 = 6x – 5x ⇒ 3,00,000 = x ⇒ x = 3,00,000 Hence, the required money is Rs 3,00,000.

25828.

Ramesh distributed his capital, half of the Capital to his wife, one third to his son and remaining 50,000/- to his daughter. Find the total amount of his capital.

Answer»

Let x be the total money.

Then, according to question,

Money given to wife = x/2

Money given to son = x/3

Money given to daughter = Rs 50,000

∴From question,

x/2 + x/3 + 50000 = x

⇒ 3x + 2x + 50,000 x 6 = 6x [Multiplying by 6]

⇒ 5x + 3,00,000 = 6x

⇒ 3,00,000 = 6x – 5x

⇒ 3,00,000 = x

⇒ x = 3,00,000

Hence, the required money is Rs 3,00,000.

Discussion

25829.	In which case and in which year do we find a sudden decrease in the IMR?
Answer» We find sudden decrease in the case of male IMR in the year 2003.

Discussion

25830.	What, according to you, may the reasons be for the female IMR being higher than the male IMR?.
Answer» I think the reason may be that the parents think that girl children are burdensome. They think it is very difficult to educate them and get them married to the males. They have to give dowry to the males. So, most of the parents don’t want female children. Hence, the female IMR is higher than the male IMR.

Discussion

25831.	What may be the reasons for the decrease in IMR rates over the years?
Answer» The reasons for the decrease in IMR rates over the years. 1. The attitude of the parents is changed with the times. 2. Increasing medical facilities. 3. Increase in the literacy rate among the girls (women). 4. Women empowerment.

25831.

What may be the reasons for the decrease in IMR rates over the years?

Answer»

The reasons for the decrease in IMR rates over the years.

1. The attitude of the parents is changed with the times.

2. Increasing medical facilities.

3. Increase in the literacy rate among the girls (women).

4. Women empowerment.

Discussion

25832.	Write the Summary of 'I Can Take Care of Myself'.
Answer» Once, a mother rat wanted to get her young daughter married to the most powerful being that she could find. She thought that the sun god was the most powerful being on earth. So, she asked the sun god if he was the most powerful being on earth. The sun god smiled and replied that the rain was greater than him as there would be no water on earth or no crop or tree without the rain. So, the mother rat asked the rain god if he was the most powerful being on earth. The rain god smiled and replied that the mountain was greater than him as he would protect the creatures. He blocked the clouds and let the water flow safely. The mother rat asked the mountain god if he was the most powerful being on earth. The mountain god smiled and replied that the worm was greater than him. He also told that the earthworm was the greatest friend of the living beings. The mother rat’s daughter came to her and asked her what she was doing. The mother replied that she was looking for the most powerful being on earth to get her married. She also told her daughter that she would be safe if she married the most powerful being on earth. The daughter replied why she would need to marry to be safe. She also told that she would need to know how to take care of herself if she wanted to be safe. To protect herself, she needed to learn to be strong and work hard. She wanted to be powerful herself, so that she could take care of herself and those that she loved. She wanted to stand on her own feet. She opined that she needed to learn more about the world and learn to live in it as a good creature. She asked for her mother’s help. She wanted her mother’s support. She was not interested in marrying anybody. She didn’t want to depend on others. She wanted to believe her power only.

Discussion

25833.	What, according to you, may the reasons be for the death of more than half of both male and female infants?
Answer» The reasons for the death of more than half of both male and female infants. 1. The superstitious beliefs of the parents. (Village parents) 2. Lack of proper medical facilities. 3. In some cases the young couples don’t want the children in the earlier days after their marriage. They think that the children will obstruct their privacy.

25833.

What, according to you, may the reasons be for the death of more than half of both male and female infants?

Answer»

The reasons for the death of more than half of both male and female infants.

1. The superstitious beliefs of the parents. (Village parents)

2. Lack of proper medical facilities.

3. In some cases the young couples don’t want the children in the earlier days after their marriage. They think that the children will obstruct their privacy.

Discussion

25834.	The value of 1- \(\frac{1}{20}+\frac1{20^2}-\frac1{20^3}\)+ ......... correct to 5 places of decimal is(a) 1.05 (b) 0.95238 (c) 0.95239 (d) 10.5
Answer» (b) 0.95238 Given exp. = 1 - \(\frac{1}{20}+\frac1{20^2}-\frac1{20^3}\) + ............ = 1 - \(\frac{1}{20}+\frac1{400}-\frac1{8000}\) + ........... = 1 – 0.05 + 0.0025 – 0.000125 +....... = 1.0025 – 0.050125 = 0.952375 = 0.95238.

25834.

The value of 1- \(\frac{1}{20}+\frac1{20^2}-\frac1{20^3}\)+ ......... correct to 5 places of decimal is(a) 1.05 (b) 0.95238 (c) 0.95239 (d) 10.5

Answer»

(b) 0.95238

Given exp.
= 1 - \(\frac{1}{20}+\frac1{20^2}-\frac1{20^3}\) + ............

= 1 - \(\frac{1}{20}+\frac1{400}-\frac1{8000}\) + ...........

= 1 – 0.05 + 0.0025 – 0.000125 +.......

= 1.0025 – 0.050125 = 0.952375 = 0.95238.

Discussion

25835.	What do you think is the most important thing to learn to live well?
Answer» The most important thing to learn to live well is that one should not depend upon others. One should take care of oneself. One should stand on one’s own feet.

Discussion

25836.	A man first sold \(\frac23\) rd of his total quantity of rice and 100 kg. Again he sold \(\frac12\) of the remaining quantity and 100 kg. If the total remaining quantity of the stock is 150 kg. Then, what was the original stock of rice ? (a) 2100 kg (b) 1800 kg (c) 2400 kg (d) 2000 kg
Answer» (b) 1800 g Let the original stock of rice be x kg. Parts of the stock sold first time = \(\bigg(\frac{2x}{3}+100\bigg)\) kg ∴ Remaining stock = \(\bigg[x-\bigg(\frac{2x}{3}+100\bigg)\bigg]\) kg = \(\big(\frac{x}{3}-100\big)\)kg Part of the stock sold second time = \(\bigg[\frac12\bigg(\frac{x}3-100\bigg)+100\bigg]\) = \(\bigg(\frac{x}{6}-50+100\bigg)\) kg = \(\big(\frac{x}{6}+50\big)\) kg ∴ Remaining stock = \(\big(\frac{x}{3}-100\big)\) - \(\big(\frac{x}{6}+50\big)\) = \(\big(\frac{x}{3}-\frac{x}6-100-50\big)\) kg = \(\big(\frac{x}{6}-50\big)\) kg Given, \(\frac{x}{6} - 150 = 150 ⇒ \frac{x}{6}=300 ⇒x=\) 1800 kg.

25836.

A man first sold \(\frac23\) rd of his total quantity of rice and 100 kg. Again he sold \(\frac12\) of the remaining quantity and 100 kg. If the total remaining quantity of the stock is 150 kg. Then, what was the original stock of rice ? (a) 2100 kg (b) 1800 kg (c) 2400 kg (d) 2000 kg

Answer»

(b) 1800 g

Let the original stock of rice be x kg.

Parts of the stock sold first time = \(\bigg(\frac{2x}{3}+100\bigg)\) kg

∴ Remaining stock = \(\bigg[x-\bigg(\frac{2x}{3}+100\bigg)\bigg]\) kg

= \(\big(\frac{x}{3}-100\big)\)kg

Part of the stock sold second time

= \(\bigg[\frac12\bigg(\frac{x}3-100\bigg)+100\bigg]\)

= \(\bigg(\frac{x}{6}-50+100\bigg)\) kg = \(\big(\frac{x}{6}+50\big)\) kg

∴ Remaining stock = \(\big(\frac{x}{3}-100\big)\) - \(\big(\frac{x}{6}+50\big)\)

= \(\big(\frac{x}{3}-\frac{x}6-100-50\big)\) kg

= \(\big(\frac{x}{6}-50\big)\) kg

Given, \(\frac{x}{6} - 150 = 150 ⇒ \frac{x}{6}=300 ⇒x=\) 1800 kg.

Discussion

25837.	A man has divided his total money in his will in such a way that half of it goes to his wife, \(\frac23\) rd of the remaining among his three sons equally and the rest among his four daughters equally. If each daughter gets Rs 20,000, how much money will each son get ?
Answer» Let the total money, the man had = Rs x Money gone to wife = Rs \(\frac{x}2\) Remaining money = \(\frac{x}2\) ∴ Sons' share = \(\frac23\) of \(\frac{x}2\) = Rs \(\frac{x}3\) Each son's share = \(\frac13\) of Rs\(\frac{x}3\) = Rs \(\frac{x}9\) Daughters' share = \(\frac{x}2\) - \(\frac{x}3\) = \(\frac{x}6\) Each daughter's share = \(\frac14\) x Rs \(\frac{x}6\) Rs \(\frac{x}{24}\) Given, \(\frac{x}{24}\) = 20000 ⇒ x = Rs 480000 ∴ Each son's share = \(\frac{480000}{9}\) = Rs 53,333.33

25837.

A man has divided his total money in his will in such a way that half of it goes to his wife, \(\frac23\) rd of the remaining among his three sons equally and the rest among his four daughters equally. If each daughter gets Rs 20,000, how much money will each son get ?

Answer»

Let the total money, the man had = Rs x

Money gone to wife = Rs \(\frac{x}2\)

Remaining money = \(\frac{x}2\)

∴ Sons' share = \(\frac23\) of \(\frac{x}2\) = Rs \(\frac{x}3\)

Each son's share = \(\frac13\) of Rs\(\frac{x}3\) = Rs \(\frac{x}9\)

Daughters' share = \(\frac{x}2\) - \(\frac{x}3\) = \(\frac{x}6\)

Each daughter's share = \(\frac14\) x Rs \(\frac{x}6\) Rs \(\frac{x}{24}\)

Given, \(\frac{x}{24}\) = 20000 ⇒ x = Rs 480000

∴ Each son's share = \(\frac{480000}{9}\) = Rs 53,333.33

Discussion

25838.	A drum can be filled with oil with a supply pipe in 40 minutes. Again the full drum can be made empty by another pipe in 60 minutes. When 2/3 part of the drum was filled, the second pipe was opened and it was stopped after 15 minutes. If the supply pipe is opened now, time required to fill the drum is(a) \(23\frac{1}{3}\) min(b) \(25\frac{2}{3}\) min(c) \(27\frac{1}{3}\) min(d) \(28\frac{2}{3}\) min
Answer» (a) \(23\frac{1}{3}\) min Part of the drum that can be emptied in 15 min. by the second pipe = 15 x \(\frac{1}{60}\) = \(\frac{1}{4}\) After 15 min, part of the drum filled with oil = \(\frac{2}{3}\) - \(\frac{1}{4}\) = \(\frac{8-3}{12}\) = \(\frac{5}{12}\) After 15 min, part of the drum emptied = 1 - \(\frac{5}{12}\) = \(\frac{7}{12}\) Given, the whole drum can be filled in 40 min. \(\therefore\) \(\frac{7}{12}\) part of the drum can be filled in \(\big(40\times\frac{7}{12}\big)\)min = \(\frac{70}{3}\) min = 23\(\frac{1}{3}\) min.

25838.

A drum can be filled with oil with a supply pipe in 40 minutes. Again the full drum can be made empty by another pipe in 60 minutes. When 2/3 part of the drum was filled, the second pipe was opened and it was stopped after 15 minutes. If the supply pipe is opened now, time required to fill the drum is(a) \(23\frac{1}{3}\) min(b) \(25\frac{2}{3}\) min(c) \(27\frac{1}{3}\) min(d) \(28\frac{2}{3}\) min

Answer»

(a) \(23\frac{1}{3}\) min

Part of the drum that can be emptied in 15 min. by the second pipe = 15 x \(\frac{1}{60}\) = \(\frac{1}{4}\)

After 15 min, part of the drum filled with oil = \(\frac{2}{3}\) - \(\frac{1}{4}\) = \(\frac{8-3}{12}\) = \(\frac{5}{12}\)

After 15 min, part of the drum emptied = 1 - \(\frac{5}{12}\) = \(\frac{7}{12}\)

Given, the whole drum can be filled in 40 min.

\(\therefore\) \(\frac{7}{12}\) part of the drum can be filled in \(\big(40\times\frac{7}{12}\big)\)min = \(\frac{70}{3}\) min = 23\(\frac{1}{3}\) min.

Discussion

25839.	Which of the following is an optional responsibility?(A) Business responsibility(B) Ethical responsibility(C) Legal responsibility(D) None of these
Answer» Correct option is (B) Ethical responsibility

Discussion

25840.	Explain ethical responsibility of business.
Answer» Ethical responsibilities: After a company has fulfilled the two big responsibilities i.e. economic and legal, a company can think of ethical responsibilities. Ethical responsibilities are responsibilities that a company needs to fulfill because its owners believes it’s the right thing to do and not that they have to do it forcefully. Ethical responsibilities could include being environment friendly, paying fair wages, participating in government programmes, etc.

25840.

Explain ethical responsibility of business.

Answer»

Ethical responsibilities:

After a company has fulfilled the two big responsibilities i.e. economic and legal, a company can think of ethical responsibilities.
Ethical responsibilities are responsibilities that a company needs to fulfill because its owners believes it’s the right thing to do and not that they have to do it forcefully. Ethical responsibilities could include being environment friendly, paying fair wages, participating in government programmes, etc.

Discussion

25841.	Explain ethical responsibility as a level of social responsibility for a business.
Answer» Ethical responsibilities are responsibilities that , a company needs to fulfill because its owners believes it’s the right thing to do and not that it has to do it forcefully.

Discussion

25842.	The fluid contained in a bucket can fill four large bottles or seven small bottles. A full large bottle is used to fill an empty small bottle. What fraction of the fluid is left over in the large bottle when the small one is full ?(a) \(\frac27\) (b) \(\frac37\) (c) \(\frac47\) (d) \(\frac57\)
Answer» Let the capacity of the bucket be x litres. Then, Capacity of 1 large bottle = \(\frac{x}4\) Capacity of 1 small bottle = \(\frac{x}7\) Fluid left in large bottle = \(\frac{x}4\) - \(\frac{x}7\) = \(\frac{3x}{28}\) ∴ Required fraction = \(\frac{\frac{3x}{28}}{\frac{x}4} = \) \(\frac{3x}{28}\) x \(\frac4x\) = \(\frac37.\)

25842.

The fluid contained in a bucket can fill four large bottles or seven small bottles. A full large bottle is used to fill an empty small bottle. What fraction of the fluid is left over in the large bottle when the small one is full ?(a) \(\frac27\) (b) \(\frac37\) (c) \(\frac47\) (d) \(\frac57\)

Answer»

Let the capacity of the bucket be x litres. Then,

Capacity of 1 large bottle = \(\frac{x}4\)

Capacity of 1 small bottle = \(\frac{x}7\)

Fluid left in large bottle = \(\frac{x}4\) - \(\frac{x}7\) = \(\frac{3x}{28}\)

∴ Required fraction = \(\frac{\frac{3x}{28}}{\frac{x}4} = \) \(\frac{3x}{28}\) x \(\frac4x\) = \(\frac37.\)

Discussion

25843.	A picture hall has seats for 820 persons. At a recent film show, one usher guessed it was 3/4 full, another that it was 2/3 full. The ticket office reported 648 sales. Which usher (first or second) made the better guess?
Answer» Correct answer is First Usher

Discussion

25844.	Which of the following processes shows how an individual seeks information about a certain issue and how he interprets that information? (a) Dogmatism (b) Perception (c) Stereotyping (d) Cognitive dissonance
Answer» Correct option is (b) Perception

Discussion

25845.	The scripture which interprets ethical behaviour of human refers to(A) Sociology(B) Psychology(C) Political science(D) Ethics
Answer» Correct option is (D) Ethics

Discussion

25846.	Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Answer» Total part = 1 Part eaten by Ritu = \(\frac{3}{5}\) Part eaten by Somu = Total part – part eaten by Ritu = 1 - \(\frac{3}{5}\) = \(\frac{5 - 3}{5} = \frac{2}{5}\) Now, We have to tell who ate the larger share so, we have to compare, Amount she ate larger by = \(\frac{3}{5}\) - \(\frac{2}{5}\) = \(\frac{3 - 2}{5} = \frac{1}{5}\) Ritu ate large share by \(\frac{1}{5}\)

25846.

Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?

Answer»

Total part = 1
Part eaten by Ritu = \(\frac{3}{5}\)
Part eaten by Somu = Total part – part eaten by Ritu

= 1 - \(\frac{3}{5}\) = \(\frac{5 - 3}{5} = \frac{2}{5}\)

Now,
We have to tell who ate the larger share so, we have to compare,

Amount she ate larger by = \(\frac{3}{5}\) - \(\frac{2}{5}\)

= \(\frac{3 - 2}{5} = \frac{1}{5}\)

Ritu ate large share by \(\frac{1}{5}\)

Discussion

25847.	Write the correct answer. The picture interprets (a) 1/4 ÷ 3(b) 3 x 1/4(c) 3/4 x 3(d) 3 ÷ 1/4
Answer» Correct answer is (b) 3 x 1/4

25847.

Write the correct answer. The picture interprets (a) 1/4 ÷ 3(b) 3 x 1/4(c) 3/4 x 3(d) 3 ÷ 1/4

Answer»

Correct answer is (b) 3 x 1/4

Discussion

25848.	Fill in the blanks to make the statements true.Rani ate 2/7 part of a cake while her brother Ravi ate 4/5 of the remaining. Part of the cake left is __________
Answer» Rani ate 2/7 part of a cake while her brother Ravi ate 4/5 of the remaining. Part of the cake left is 1/7

Discussion

25849.	Identify the principal of management in the following lines. “Therefore after six months when the business was doing well, she awarded a cash bonus to each of these employees to honour her commitment. (a) Discipline (b) Initiative (c) Scalar chain (d) Remuneration
Answer» Correct option is (a) Discipline

Discussion

25850.	Management is an _______ (a) Art (b) Science (c) Art and Science (d) Art or Science
Answer» Management is an Art and Science.

Discussion

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