WBBSE Class 10 Model Question Paper 2023 Mathematics set 2

Model Question Paper 2023 Mathematics set 2

 

Model Question Paper 2023 Mathematics set 2 MCQs

 

Question 1. Sum of the two roots of equation 5x² – 2x + 1 = 0 is

1. 2
2. 1
3. 1/2
4. 2/5

Solution:

Given

5x² – 2x + 1 = 0

Sum of the two roots of equation = -b/a

= -(-2/5)

= 2/5

Answer. 4. 2/5

Question 2. Samir invests Rs. 4,000 for 3 months and Amita invests Rs. 3,000 for 5 months in a business. Profit will be distributed in the ratio

1. 4:3
2. 3:4
3. 4:5
4. 5:4

Solution:

Given

Samir invests Rs. 4,000 for 3 months and Amita invests Rs. 3,000 for 5 months in a business.

Ratio of profit = 4000 x 3: 3000 x 5 = 12:15=4:5

Answer. 3. 4:5

Question 3. x x 1/y and y = 2/5 when x = 5; x = 1/6 the value of y is

1. 1/3
2. 6
3. 12
4. 18

Solution : x = k/t or,5 = k 5/2

.. k = 2

Y = k/x = 2/½

= 12

The value of y is = 12

Answer. 3. 12

Question 4. In the adjoining figure, AB is the diameter of a circle with center O. If ZADC = 140° and /CAB = x°, the value of x is

1. 80°
2. 40°
3. 50°
4. 30°

Solution: CAB = x.x + 40° +90° = 180°

∴ x = 50°

Answer. 3. 50°

 

Question 5. If the length of the tower and the length of the shadow of a stick having a length of 20 meters are 50 meters and 10 meters respectively, the length of the tower is

1. 120 m.
2. 250 m.
3. 25 m.
4. 100 m.

Answer. 4. 100 m.

Given

If the length of the tower and the length of the shadow of a stick having a length of 20 meters are 50 meters and 10 meters respectively,

The length of the tower is 100 m.

“WBBSE Class 10 set 2 Mathematics model paper with solutions 2023”

Question 6. If the height of a solid right circular cone is 15 cm and the length of the diameter of the base is 16 cm, then the area of the lateral surface is

1. 120 sq cm.
2. 240π sq cm.
3. 136π sq cm.
4. 130π sq cm.

Solution: лгl= л x 8 x 17 sq cm

If the height of a solid right circular cone is 15 cm and the length of the diameter of the base is 16 cm

= 136л sq cm.

The area of the lateral surface is  130 sq cm.

Answer. 4. 130 sq cm.

Question 7. If the total surface area of a solid hemisphere is 1477, the length of the radius of the base of the hemisphere is

1. 14 cm.
2. 7 cm.
3. 21 cm.
4. 7.5 cm.

Solution: 3r2 = 147

If the total surface area of a solid hemisphere is 1477

∴ r=7

Answer. 2. 7 cm.

The length of the radius of the base of the hemisphere is 2. 7 cm.

 

Question 2. Answer the following questions :

1. If Α α 1/C and C α 1/B, find the variation relation between A and B.

Solution: A α 1/C       .. A= K₁ /C

C α 1/B    .. C = K₂/B     (K₁ & K₂ are constant)

A = K₁/K₂B     ∴ Aα B.

2. Fill in the blanks

1. If a straight line intersects the circle at the two points, this straight line is called the Intercept of the circle.

2. The length of the radii of the two circles are 4 cm and 5 cm. If two circles touch each other externally, the distance between the two centers is  9 cm.

“West Bengal Class 10 Mathematics set 2 model question paper 2023 with answers”

3. If the surface area of a sphere is S and the volume is V, write the value of S³/V².

Solution:  S³/V²= 64л³r⁶/ 64/27л²r⁶

= 27π

Answer: 27π.

 

4. The lateral surface area of ae is √5 times its base area. Find the ratio between the height of the cone and the length ctius of its base.

Solution: 

Answer: 2/1

Question 3. Shova, Masud, and Rabeya started a business with a capital of Rs. 3,000, Rs. 3500, and Rs. 2500 respectively. They have decided to divide part 1/3 of the total profit equally and the remaining profit in the ratio of their capital invested. What will be the share of profit of each of them if the amount of profit at the end of the year is Rs. 810? 

Solution:

Shova, Masud, and Rabeya started a business with a capital of Rs. 3,000, Rs. 3500, and Rs. 2500 respectively. They have decided to divide part 1/3 of the total profit equally and the remaining profit in the ratio of their capital invested.

The ratio of capitals of Shova, Masud & Rabeya are

3000 :3500 :2500 = 30: 35: 25=6:7:5

Total Profit = Rs. 810

profit = Rs. 810 x 1/3

= Rs. 270.

Of Rs. 270, each will get = Rs. 270 ÷ 3 = Rs. 90. 

The rest of the profit= is Rs. (810270) = Rs. 540.

Out of Rs. 540,

Shova will get = 6/6+7+5 x Rs. 540 

= 6/18 × 540 

= Rs. 180

Masud will get= 7/18 x Rs. 540 

= Rs. 210

& Rabeya will get = 5/18 x Rs. 540 

= Rs. 150

∴ Shova will get = Rs. (90+ 180) 

= Rs. 270

Masud will get in total = Rs. (90+210) 

= Rs. 300

& Rabeya will get in total = Rs. (90+150) 

= Rs. 240

Or, Shakil and Mohuya jointly started a business with a capital of Rs. 30,000 and Rs. 50,000 respectively. After 6 months Shakil invested Rs. 40,000 more in the business, but Mohuya withdrew Rs. 10,000 for personal needs. If the profit amount is Rs. 19,000 at the end of the year, how many shares of profit each of them should get?

Solution:

Given

Shakil and Mohuya jointly started a business with a capital of Rs. 30,000 and Rs. 50,000 respectively. After 6 months Shakil invested Rs. 40,000 more in the business, but Mohuya withdrew Rs. 10,000 for personal needs. If the profit amount is Rs. 19,000 at the end of the year

Ratio of Capitals of Shakil & Mohuya throughout the year

= Rs. (30,000 12+ 40000 6): Rs. (50,000 6+ 40,000 6)

= Rs. (3,60,000+ 2,40,000): Rs. (3,00,000+ 2,40,000)

= 6,00,000 5,40,000 

= 60 :54 = 10:9

Total profit= Rs. 19,000

Shakil will get = 10/10+9 Rs. 19,000 

= Rs. 10/19 19,000 

= Rs. 10,000. Ans.

Mahna will get = 9/10+9 Rs. 19,000 

= Rs. 9/19 19,000 

= Rs. 9,000. Ans.

Question 4. If a c b and b c c, show that a³b³+ b³c³+ c³a³ α abc(a³ + b³+ c³).

Solution: aα b or, a =K₁b

Or, Solve: x-2 / x+2 +6 (x-2 / x+6) = 1 ,(x ≠ -2,6)

Solution : x-2 / x+2 +6 (x-2 / x+6) = 1

 

“WBBSE 2023 Class 10 Mathematics model set 2 question paper with step-by-step solutions”

Question 5. Prove that if two circles touch each other externally, then the point of contact will lie on the line joining the two centers.

Solution: 

Given: Two circles with centers A and B touch each other at a point P. To prove A, P, and B are collinear.

Construction: A, P, and B, P are joined

Proof: Two circles with centers A and B touch each other at a point P.

∴ There exists a common tangent at point P.

Let ST is the common tangent, which touches both circles at point P.

∴ ST is the tangent of a circle with center A and AP is the radius through the point of contact

∴ AP ⊥ ST

Again, ST is tangent to the circle with center B and BP is the radius through the point of contact.

∴ BP ⊥ ST

∴ AP and BP are both perpendicular to ST at some point P.

∴ AP and BP lie on the same straight line, i.e., A, P, and B are collinear.

 

 

Or, Prove that if a perpendicular is drawn from the vertex containing the right angle of a right-angled triangle on the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and are similar to each other. 

Solution: ABC is a right-angled triangle whose∠A is a right angle and from right angular point A, AD is perpendicular on the hypotenuse BC.

To prove:

1. ΔDBA and ΔABC are similar to each other.

2. ΔDAC and ΔABC are similar to each other.

3. ΔDBA and ΔDAC are similar to each other.

Proof: In ΔDBA and ΔABC,

∠BDA = <BAC = 90° and ∠ABD = ∠CBA. So remaining

∠BAD = ∠BCA

∴ ΔDBA and ΔABC are similar to each other.

[1 is proved]

Again, in ΔDAC and ΔABC,

∠ADC= <BAC = 90°

∠ACD = ∠BCA. So remaining ∠CAD =∠CBA

∴ ΔDAC and ΔABC are similar to each other.

[2 is proved]

ΔDBA and ΔABC are similar to each other.

Again, ΔDAC and ΔABC are similar to each other.

So, ΔDBA and ΔDAC are similar to each other.

[3 is proved]

 

Question 6. BAC=1 right angle of a right-angled triangle ABC. AD is perpendicular to ΔABC BC² hypotenuse BC. Prove that ΔΑΒC/ΔACD = BC²/AC².

Solution: In right-angled AABC, A = 90°, & AD is perpendicular on hypotenuse BC.

To prove: ΔΑΒC/ΔACD = BC²/AC².

Proof: ACD & ABC are equiangular.

 

Question 7. Draw a triangle of which the length of one side is 6.8 cm and measures of two angles adjacent to this side are 60° and 75° and draw the circumcircle of this triangle.

Solution: The constructed triangle is ABC whose AB = 6.8 cm,

BAC = 60° and ABC = 75°

The center of the circumcircle.

 

 

Or, Draw a circle with center O having a radius of 2.5 cm in length. Take a point P outside the circle at a distance of 5 cm. from ‘O’. Draw two tangents to the circle from point P.

Solution: In AP is a tangent to the circle with center O, from an external point A.

 

 

Question 8. Three spheres made of copper with radii of 3 cm, 4 cm, and 5 cm are melted and a large sphere is made. What is the length of the radius of the large sphere?

Solution:

Given

Three spheres made of copper with radii of 3 cm, 4 cm, and 5 cm are melted and a large sphere is made.

Volume of 1st sphere with radius 3 cm = π (3)³ cu cm.

The volume of the 2nd sphere with a radius of 4 cm = 4/3 π (4)³ cu cm.

& Volume of 3rd sphere with radius 5 cm = 4/3 π (5) cu cm.

Total volume of 3 small spheres = 4/3  π (3)³+ 4/3  π (4)³ + 4/3  π (5)³

= π(33 +43 +53) cu cm.

= π(27 +64 +125) cu.cm.

= 4/3 π x 216 cu cm.

Now, let the radius of the new big sphere = R cm.

Its volume = 4/3 πR³

According to the problem,

4/3πr³ = 4/3 π x 216

∴ R3 = 216

∴ R = 3√216

= 6 cm.

The length of the radius of the large sphere = 6 cm.

“Class 10 WBBSE 2023 model question paper Mathematics set 2 with detailed answers”

Or, If the length of the radius of the base is 12 meters and the height is 5 meters, what will be the cost to make a tent in the shape of the right circular cone at the rate of Rs. 3.50 per sq meter?

Solution:

Given

If the length of the radius of the base is 12 meters and the height is 5 meters

Here radius (r) = 12 m & height (h) = 5 m.

∴ Slant height (l) = √r²+h²

= √12² +5²

= √144+25

= √169

= 13 m.

Curved Surface area of cone

πrl= 22/7 x 12 x 13 sqm

= 3432/7 sqm.

∴ The total cost of making the tent at the rate of Rs. 3.5/sqm.

= Rs. 3.5 x 3432/7

= Rs. 1716. Ans.