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QUESTION: 1

A hollow sphere of internal and external radius 3 cm and 5 cm respectively is melted into a solid right circular cone of diameter 8 cm. The height of the cone is

Solution:

We know that the formula of the volume of a hollow sphere is

4π(R^{3} – r^{3})

Here, R = external radius and r = internal radius

A hollow sphere of internal and external radius 3 cm and 5 cm respectively

So the volume of the hollow sphere = [4×π×(5^{3} – 3^{3})]/3 cc

Now, after melting this sphere, we will get a right circular cone, which’s diameter is 8 cm

So, radius of that cone = 8/2 cm = 4 cm

We know that the formula of the volume of a right circular cone is πr^{2}h/3

Here, r is the radius of the cone and h is the height of the cone

From the question,

we can make the equation,

π × 4^{2} × h/3 = [4×π×(5^{3} – 3^{3})]/3

⇒ 4h = 98

⇒ h = 24.5

So, the height of the cone is 24.5 cm

QUESTION: 2

A and B can together do a piece of work in 28 days. If A, B and C can together finish the work in 14 days, how long will C take to do the work by himself?

Solution:

If A and B can together do a piece of work in 28 days which means in 1 day, A and B will finish 1/28^{th} of the work.

If A, B and C can together finish the work in 14 days which means in 1 day, A, B and C will finish 1/14^{th} of the work.

Putting the first value in second

QUESTION: 3

Two cars are moving with speeds v_{1}, v_{2} towards a crossing along two roads. If their distances from the crossing be 40 metres and 50 metres at an instant of time then they do not collide if their speeds are such that

Solution:

then they will collide i.e. cars will reach at the same time.

QUESTION: 4

2/5 Part of a mixture of 3 l is water and the rest is sugar syrup. Another mixture of 8 l contains 3/5 part water and the rest is sugar syrup. It both mixture are mixed together then what will be the ratio of water & sugar syrup in the new mixture?

Solution:

Assume,

1) Mixture 1: mixture of 2/5 part water & rest sugar syrup.

2) Mixture 2: mixture of 3/5 part water & rest sugar syrup.

3) Mixture 3: mixture of mixture 1 & mixture 2.

Amount of water and sugar syrup in different mixtures are represented in tabulated form for better understanding.

∴ The ratio of water & sugar syrup in the new mixture = 6:5

QUESTION: 5

A wire is bent in the form of an equilateral triangle encloses a region having area of 121√3 cm². If the same wire is rebent in the form of a circle, its radius will be:

Solution:

We have area of an equilateral triangle = (√3/4) × side × side

⇒ 121√3 = (√3/4) × side × side

⇒ side × side = 121 × 4

⇒ side = 22

Perimeter = 3 × side

= 3 × 22

= 66 cm

We now have a circle with perimeter 66 cm.

⇒ 2 × (22/7) × radius = 66 cm

⇒ (44/7) × radius = 66 cm

⇒ Radius = 7 × 1.5

= 10.5 cm

QUESTION: 6

In a class of 50 students, 20 students had an average score of 70 and remaining had an average score of 50. What is the average score of the class?

Solution:

Total number of students = 50

20 students had an average score of 70

∴ Total score of these 20 students

= Average score × Number of students

= 70 × 20 = 1400

Remaining number of students = 50 – 20 = 30

Remainingstudents had an average score of 50

∴ Total score of these 30 students

= Average score × Number of students

= 50 × 30 = 1500

∴ Sum of scoresof these 50 students = 1400 + 1500 = 2900

∴ Average score of these 50 students

= Sum total of these 50 students/50 = 2900/50 = 58

QUESTION: 7

In a factory 60% of the workers are above 30 years and of these 75% are males and the rest are females. If there are 1350 male workers above 30 years, the total number of workers in the factory is

Solution:

Let the total number of workers in the factory be X. Then,

Total number of workers above 30 years of age

Also it is given that out of these 75% are male. Thus

Total number of male workers above 30 years of age

It is given that total number of male workers above 30 years

= 1350

Hence,

⇒ 0.45X = 1350

⇒ X = 3000

∴ Total workers in the factory are 3000.

QUESTION: 8

The difference between simple interest and compound interest for 2 years on the sum Rs. 2900 at a certain rate is Rs. 14.21. What is annual rate of interest?

Solution:

We know that the difference between the simple interest (SI_{2}) & compound interest (CI_{2}) for 2 years [compounded annually] on a sum of Rs. P at a rate R is,

Now the given information, P = Rs. 2900, CI_{2} – SI_{2} = Rs. 14.21

Putting the values we get,

The annual rate of interest is 7%

QUESTION: 9

The third proportional to

Solution:

Let, the third proportional is ‘x’

QUESTION: 10

ABC is a right angled triangle, B being the right angle. Midpoints of BC and AC are respectively B’ and A’. The ratio of the area of the quadrilateral AA’ B’B to the area of the triangle ABC is

Solution:

Formulas:

Area of a triangle

Area of a trapezium =

Triangle ABC and A’B’C are similar by SAS

By property of similar triangles,

∠B = ∠A’B’C and 2A’B’ = AB

∴ A’B’ is parallel to AB.

Thus quadrilateral AA’ B’B is a trapezium.

In quadrilateral AA’ B’B,

Height = BB’

⇒ Height = BC/2

Sum of parallel sides = (A’B’ + AB) = 3AB/2

Area of quadrilateral AA’ B’B

⇒ Area of quadrilateral AA’ B’B

Area of triangle ABC

Ratio area of quadrilateral AA’ B’B to area of triangle ABC

⇒ Ratio of area of quadrilateral AA’ B’B to area of triangle ABC = 3 : 4

QUESTION: 11

The cost of making an article is divided between materials, labor and overheads in the ratio 3 : 4 : 2. If the cost of material is Rs.33.60 the cost of article is

Solution:

Let the multiplying factor be ‘x’.

Materials = 3x, Labor = 4x and Overheads = 2x

Given that, materials cost = Rs.33.60

As per sum,

⇒ 33.60 = 3x

⇒ x = 11.20

⇒ Labour = 4x = 4(11.20) = 44.80

⇒ Overheads = 2x = 2(11.20) = 22.40

∴ Cost of article = 33.60 + 44.80 + 22.40 = 100.80.

QUESTION: 12

If a = √2 + 1, b = √2 – 1, then the value of

Solution:

Given,

a = √2 + 1

b = √2 – 1

Given expression is,

= 2/2

= 1

QUESTION: 13

If ‘a’ and ‘b’ are two odd positive integers, by which of the following is (a^{4} – b^{4}) always divisible?

Solution:

We know that, Summation and Subtraction between two odd integers gives an even integer.

Given, a and b are two odd positive integers.

∴ a + b = 2k_{1} ……(1)

And a – b = 2k_{2 }........(2)

Where, k_{1} and k_{2}are two integers.

(1) + (2) ⇒ 2a = 2(k_{1} + k_{2}) ⇒ a = k_{1} + k_{2}

(1) - (2) ⇒ 2b = 2(k_{1} - k_{2}) ⇒ b = k_{1} - k_{2}

∴ a^{2} + b^{2}

= (k_{1} + k_{2})^{2} + (k_{1} - k_{2})^{2}

= k_{1}^{2} + k_{2}^{2} + 2k_{1}k_{2} + k_{1}^{2} + k_{2}^{2} - 2k_{1}k_{2}

= 2(k_{1}^{2} + k_{2}^{2}) ………..(3)

Given expression is,

a^{4} – b^{4}

= (a^{2} – b^{2})(a^{2} + b^{2})

= (a + b) (a - b)(a^{2} + b^{2})

Putting values from (1), (2) and (3)

= 2k_{1 }× 2k_{2} × 2(k_{1}^{2} + k_{2}^{2})

= 8× k_{1 }× k_{2} × (k_{1}^{2} + k_{2}^{2})

∴ a^{4} – b^{4 }= 8× k_{1 }× k_{2} × (k_{1}^{2} + k_{2}^{2}) is always divisible by 8.

QUESTION: 14

If (x - 2) and (x + 3) are the factors of x^{2} + k_{1}x + k_{2} then

Solution:

If (x - 2) and (x + 3) are the factors of the equation, then that means that

⇒ x – 2 = 0

⇒ x = 2

Similarly,

⇒ x + 3 = 0

⇒ x = -3

Putting the value of one of the x in the equation such that the value is zero.

⇒ x^{2} + k_{1}x + k_{2}

Putting the value of ‘x’ as 2, we get

⇒ (2)^{2} + k_{1}(2) + k_{2 }= 0

⇒ 4 +2k_{1}+ k_{2 }= 0

Putting the value of ‘x’ as -3, we get

⇒ (-3)^{2} + k_{1}(-3) + k_{2 }= 0

⇒ 9 -3k_{1}+ k_{2 }= 0

Subtracting both the equations:

⇒ 4 +2 k_{1}+ k_{2 }- (9 -3 k_{1}+ k_{2}) = 0

⇒ 4 +2 k_{1}+ k_{2 }- 9 + 3 k_{1}- k_{2 }= 0

⇒ 5 k_{1 }- 5 = 0

⇒ 5 k_{1 }= 5

⇒ k_{1 }= 1

Putting the value of k_{1 }= 1,

⇒ 4 +2 k_{1}+ k_{2 }= 0

⇒ 4 +2 (1)+ k_{2 }= 0

⇒ 4 +2 + k_{2 }= 0

⇒ 6 + k_{2 }= 0

⇒ k_{2 }= -6

Alternate method:-

(x – 2) and (x + 3) are factors of x^{2} + k_{1}x + k_{2}

∴ (x – 2)(x + 3) = x^{2} + k_{1}x + k_{2}

⇒ x^{2} + x – 6 = x^{2} + k_{1}x + k_{2}

Comparing, we get, k_{1} = 1 and k_{2} = - 6

QUESTION: 15

If ab + bc + ca = 0, then the value of

Solution:

In the given question let us add (ab + bc + ca) to all the denominators, such that the equation becomes:

Taking L.C.M of the above equation

⇒ 0

QUESTION: 16

The heights of two poles are 180 m and 60 m respectively. If the angle of elevation of the top of the first pole form the foot of the second pole is 60°, what is the angle of elevation of the top of the second pole form the foot of the first?

Solution:

Given, heights of two poles are 180 m and 60 m.

We know that the distance between the two poles acts as base and will be same so,

For the first pole,

tan 60° = height/base

⇒ √3 = 180m / b

⇒ b = 60√3

tan θ = h / 60√3

⇒ tan θ = 1/√3

⇒ θ = 30°

QUESTION: 17

If 3 sin θ + 5 cos θ = 5, then the value of 5 sin θ – 3 cos θ wil be

Solution:

Formula-

sin^{2}θ + cos^{2} θ = 1

Given,

3 sin θ + 5 cos θ = 5 …….eq(1)

Let, 5 sin θ – 3 cos θ = x …..eq(2)

On squaring and adding both equations

(3 sin θ + 5 cos θ)^{2}+ (5 sin θ – 3 cos θ)^{2} = 25 +x^{2}

⇒ 9sin^{2 }θ + 25 cos^{2 }θ +30sinθ cosθ + 25 sin^{2 }θ + 9cos^{2}θ – 30sinθ cosθ = 25 +x^{2}

⇒ 9sin^{2 }θ + 9cos^{2 }θ+ 25 cos^{2 }θ +25 sin^{2 }θ = 25 +x^{2}

⇒ 9(sin^{2 }θ +cos^{2 }θ) + 25(sin^{2 }θ +cos^{2 }θ) =25 +x^{2}

⇒ 9 + 25 = 25 +x^{2}

⇒ x^{2}= 9

⇒ x = ±3

QUESTION: 18

A shopkeeper sold an item for Rs. 1,510 after giving a discount of and there by incurred a loss of 10%. Had he sold the item without discount, his net profit would have been

Solution:

Given, selling price (SP) = 1510,

Marked price – 49/2% of marked price = 1510

⇒ Marked price = 1510/0.755 = Rs. 2000

Marked price is the selling price without discount.

Given, there is a loss of 10%.

QUESTION: 19

If A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram taken in order, then find the values of x and y.

Solution:

Given, vertices of parallelogram are A(1, 2), B(4, y), C(x, 6) and D(3, 5)

We know that diagonals of a parallelogram bisect each other, so let the diagonals cross at point O.

Since O divides AC and BD in 1:1 ratio so,

Coordinate of O (w.r.t. AC) =

Coordinate of O (w.r.t. BD)

Since both the coordinate s are same so equating them we get,

1 + x = 7

⇒ x = 6

And,

8 = 5 + y

⇒ y = 3

QUESTION: 20

In a circle of radius 5 cm, AB and AC are two equal chords such that AB = AC = 6 cm. The length of BC (in cm) is

Solution:

Pythagoras theorem:-

Hypotenuse^{2} = perpendicular^{2} + base^{2}

(a - b)^{2} = a^{2} + b^{2} - 2ab

Let O be the center of the circle.

Since Δ ABC is isosceles therefore line AO will bisect BC perpendicularly.

So, PB = CP

By Pythagoras theorem,

AB^{2} = AP^{2} + BP^{2}

⇒ 36 = AP^{2} + BP^{2} …(1)

And,

BO^{2} = BP^{2} + OP^{2}

⇒ 25 = BP^{2} + (5 – AP)^{2} [∵AP + OP = 5]

⇒ 25 = BP^{2} + 25 + AP^{2} – 10AP …(2)

Substituting value of AP^{2 }+ BP^{2} in (2)

⇒ 10AP = 36

⇒ AP = 3.6 cm

⇒ BP^{2} = 36 – 12.96

⇒ BP^{2} = 23.04

⇒ BP = 4.8 cm

⇒ BC = 9.6 cm

QUESTION: 21

In a Δ PQR, ∠RPQ = 90°, PR = 8 cm and PQ = 6 cm, then the radius of the circumcircle of Δ PQR is

Solution:

Pythagoras theorem,

Hypotenuse^{2} = perpendicular^{2} + base^{2}

Area of a triangle = abc/4R

Where, a, b and c are sides of the triangle and R is the radius of circumcircle.

Area of the triangle PQR = ½ × PQ × PR

Where PR = 8cm, PQ = 6cm

Area of triangle =12×8×6=24cm2=12×8×6=24cm2

Now, using Pythagoras theorem, a^{2 }+ b^{2} = c^{2}

⇒ c^{2 }= (8)^{2} + (6)^{2} = 64 + 36 = 100 = (10)^{2}

⇒ c = 10

Radius

QUESTION: 22

**Directions:** Study the following table carefully and answer the questions.

** Q. **If farmer D and farmer E, both sell 240 kg of Bajara each, what would be the respective ratio of their earnings?

Solution:

From the table,

Price per kg of Bajra sold by farmer D = 28

Price per kg of Bajra sold by farmer E = 30

Earning on 240 kg Bajra by D = 240 × 28 = Rs. 6720

Earning on 240 kg Bajra by E = 240 × 30 = Rs. 7200

∴ Required ratio

= 6720 : 7200 = 14 : 15

QUESTION: 23

**Directions:** Study the following table carefully and answer the questions.

** Q. **What is the average price per kg of Bajra sold by all the farmers together?

Solution:

Price per kg of Bajra sold by farmer A = 22

Price per kg of Bajra sold by farmer B = 24.5

Price per kg of Bajra sold by farmer C = 21

Price per kg of Bajra sold by farmer D = 28

Price per kg of Bajra sold by farmer E = 30

∴ Average price per kg of Bajra sold by all the farmers together

= (Sum of price per kg of Bajra)/number of farmer

= (22 + 24.5 + 21 + 28 + 30)/5

= 125.5/5 = 25.1

QUESTION: 24

**Directions:** Study the following table carefully and answer the questions.

** Q. **If farmer A sells 350 kg of rice, 150 kg of corn and 250 kg of jowar, how much would he earn?

Solution:

**From the table,**

Price per kg of rice sold by farmer A = 30

Price per kg of corn sold by farmer A = 22.5

Price per kg of jowar sold by farmer A = 18

∴ Earning on 350 kg of rice by farmer A = 350 × 30 = Rs. 10500

Earning on 150 kg of corn by farmer A = 150 × 22.5 = Rs. 3375

Earning on 250 kg of jowar by farmer A = 250 × 18 = Rs. 4500

∴ Total earn = (10500 + 3375 + 4500) = Rs. 18375

QUESTION: 25

**Directions:** Study the following table carefully and answer the questions.

Q. Earnings on 150 kg of paddy sold by farmer B are approximately what percent of the earnings on the same amount of rice sold by the same farmer?

Solution:

**Given,**

Price per kg of paddy sold by farmer B = 25

Price per kg of rice sold by farmer B = 36

∴ Earning on 150 kg of paddy by farmer B = 150 × 25 = Rs. 3750

Earning on 150 kg of rice by farmer B = 150 × 36 = Rs. 5400

∴ Required per cent

= (3750/5400) × 100

= 3750/54 = 69.44%

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