Marksparks .arkansas

Arithmetic

• Chapter 1 Simple Interest Multiple Choice Questions
• Chapter 2 Compound Interest And Uniform Rate Of Increase Or Decrease Multiple Choice Questions
• Chapter 3 Partnership Business Multiple Choice Questions

Algebra

• Chapter 1 Quadratic Equations With One Variable Multiple Choice Questions
• Chapter 2 Ratio And Proportion Multiple Choice Questions
• Chapter 3 Quadratic Surds Multiple Choice Questions
• Chapter 4 Variation Multiple Choice Questions

Geometry

• Chapter 1 Theorems Related To Circle Multiple Choice Questions
• Chapter 2 Theorems Related To Tangent Of A Circle Multiple Choice Questions
• Chapter 3 Similarity Multiple Choice Questions
• Chapter 4 Pythagoras Theorem Multiple Choice Questions

Mensuration

• Chapter 1 Rectangular Parallelopiped Or Cuboid Multiple Choice Questions
• Chapter 2 Right Circular Cylinder Multiple Choice Questions
• Chapter 3 Right Circular Cone Multiple Choice Questions
• Chapter 4 Sphere Multiple Choice Questions

Trigonometry

• Chapter 1 Concept Of Measurement Of Angle Multiple Choice Questions
• Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions
• Chapter 3 Trigonometric Ratios Of Complementary Multiple Choice Questions
• Chapter 4 Application Of Trigonometric Ratios Height And Distance Multiple Choice Questions

Statistics

• Chapter 1 Mean Median Ogive Mode Multiple Choice Questions

WBBSE Class 10 Maths Mensuration Chapter 4 Sphere Multiple Choice Questions

Example 1. The volume of a solid sphere having a radius of 2r units length is

1. \(\frac{32}{3} \pi r^3 \text { cu.u. }\)
2. \(\frac{16 \pi r^3}{3} \text { cu.u }\)
3. \(\frac{8 \pi r^3}{3} \mathrm{cu} .\)
4. \(\frac{64 \pi r^3}{3} \text { cu.u }\)

Solution: Volume = \(\frac{4}{3} \pi(2 r)^3 \text { cu unit }=\frac{32 \pi r^3}{3} \text { cu.u }\)

∴ Answer is 1. \(\frac{32}{3} \pi r^3 \text { cu.u. }\)

Example 2. If the ratio of the volumes of two solid sphere is 1: 8, the ratio of their curved surface area is

1. 1: 2
2. 1: 4
3. 1: 8
4. 1: 16

Solution: Ratio of volume = \(\frac{\frac{4}{3} \pi}{\frac{4}{3} \pi}\left(\frac{r_1^2}{r_1^2}\right)^3=\left(\frac{1}{8}\right)\)

∴ \(\frac{r_1}{r_2}=\frac{1}{2}\)

Ratio of curved surface = \(\frac{4 \pi}{4 \pi}\left(\frac{r_1}{r_2}\right)^2=\frac{1}{4}\)

∴ Answer is 2. 1: 4

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 3. The whole surface area of a solid hemisphere with length of 7 cm radius is

1. 588 π sq. m
2. 392 π sq. m
3. 147 π sq. m
4. 98 π sq. m

Solution: Whole surface area = 3π (7)² sq. cm = 147 sq. cm

∴ Answer is 3. 147 π sq. m

Class 10 Maths Mensuration Chapter 4 Sphere MCQs

Example 4. If the ratio of curved surface areas of two solid sphere is 16: 9, the ratio of their volumes is

1. 64: 27
2. 4: 3
3. 27: 64
4. 3: 4

Solution: \(\frac{4 \pi}{4 \pi}\left(\frac{r_1}{r_2}\right)^2=\frac{16}{7}\)

or, \(\frac{r_1}{r_2}=\frac{4}{3}\)

∴ \(\frac{\frac{4}{3} \pi}{\frac{4}{3} \pi}\left(\frac{r_1}{r_2}\right)^3=\left(\frac{4}{3}\right)^3\) = 64: 27

∴ Answer is 1. 64: 27

Example 5. If numerical value of curved surface area of a solid sphere is three times of its volume the length of radius is

1. 1 unit
2. 2 unit
3. 3 unit
4. 4 unit

Solution: \(4 \pi r^2=3 \frac{4}{3} \pi r^3\) or, r = 1

∴ Answer is 1. 1 unit

Sphere Mcqs Class 10

Example 6. Volume of a sphere of radius \(\frac{r}{2}\) units will be

1. \(\frac{1}{6} \pi r^3 \text { cu. u. }\)
2. \(\frac{4}{3} \pi r^3 \text { cu. u. }\)
3. \(\frac{2}{3} \pi r^3 \text { cu. u. }\)
4. \(\frac{1}{3} \pi r^3 \text { cu. u. }\)

Solution: Volume = \(\frac{4}{3} \pi\left(\frac{r}{2}\right)^3\) cu. unit

= \(\frac{\pi r^3}{6}\) cu units

∴ Answer is 1. \(\frac{1}{6} \pi r^3 \text { cu. u. }\)

Example 7. If numerical value of curved surface area of a solid sphere Is equal to its volume, then the length of its radius is

1. 6 unit
2. 5 unit
3. 4 unit
4. 3 unit

Solution: \(4 \pi r^2=\frac{4}{3} \pi r^3\) or, r = 3

∴ Answer is 4. 3 unit

Example 8. The ratio of whole surface area of a solid sphere and solid hemisphere of same radius will be

1. 4: 3
2. 3: 4
3. 1: 2
4. None of these

Solution: Ratio = 4π (r)² : 3πr² = 4:3

∴ Answer is 1. 4: 3

Example 9. The ratio of numerical values of volume and whole surface area of a sphere is 2: 1. Numerical value of radius is

1. 5
2. 6
3. 3
4. 1

Solution: \(\frac{\frac{4}{3} \pi r^3}{4 \pi r^2}=\frac{2}{1}\)

or, \(\frac{r}{2}\) = 2, r = 6

∴ Answer is 2. 6

Class 10 Mensuration Sphere Mcqs With Answers

Example 10. Volume of a sphere is \(\frac{4}{3} \pi r^3\) πr³ cu. units. The sphere is inscribed (in) a cube. The ratio of volumes of the cube and the sphere is

1. 3: π
2. 4: π
3. 6: π
4. 8: π

Solution: Length of the side of the cube = 2r units

Ratio = (2r)³ : \(\frac{4}{3}\) πr³ = 6 : n

∴ Answer is 3. 6:

WBBSE Class 10 Maths Arithmetic Chapter 1 Simple Interest Multiple Choice Questions

Example 1. If the interest of t P at the rate of simple interest of ₹ P per annum in t years is I, then

1. I = prt
2. prtI = 100
3. prt = 100 x I
4. None of these

Solution: I=prt100 or, prt = 100 x I

∴ The answer is prt = 100 x I

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 2. A principal becomes twice its amount in 20 years at a certain rate of simple interest at the same rate of simple interest, that principal becomes thrice its amount in

1. 30 years
2. 35 years
3. 40 years
4. 45 years

Solution:

∴ The answer is 3. 40 years

Class 10 Maths Arithmetic Chapter 1 MCQs

Example 3. If a principal becomes twice its amount in 10 years the rate of simple interest per annum is

1. 5%
2. 10%
3. 15%
4. 20%

Solution:

∴ The answer is 2. 10%

Example 4. If the total interest becomes ₹ x for any principal having the rate of simple interest of x%. per annum for x years then the principal will be

1. x
2. 100x
3. 100x
4. 100x2

Solution: x = P×x×x100 or, P = 100x

∴ The answer is 3. 100x

Example 5. The total interest of a principal in n years at the rate of simple interest of r% per annum is pnr25, the principal will be

1. ₹ 2p
2. ₹ p²
3. ₹ 4p
4. ₹ p4

Solution: P×x×x100 or, principal = 4p

∴ Answer is p2

Example 6. The time in which the ratio of interest and principal be 3: 2 when rate is 10% per annum

1. 12 years
2. 15 years
3. 16 years
4. 18 years

Solution: Simple interest = PRT100

or, T = 15

∴ The answer is 15 years.

Example 7. If a certain sum of money is tripled itself in 1312 years, then the rate of interest is

1. 10%
2. 12 12%
3. 15%
4. 20%

Solution: Simple interest = PRT100

∴ The answer is 15%

Arithmetic Progression Mcqs Class 10

Example 8. If n be the no. of years, r% be the rate of interest, and the simple intcrest= Pnr50. The principal is

1. P
2. 2P
3. P2
4. None of these

Solution:

∴ Principle = 2P

∴ Answer is 2P

Example 9. A sum of money trebles itself in 30 years. Rate of interest is

1. 623
2. 15%
3. 10%
4. 6%

Solution: (3P- P) = P×R×30100

∴ Answer is 623

Wbbse Class 10 Maths Arithmetic Notes

Example 10. If the interest of P in n years at r% Simple interest be I then

1. I = Prn
2. Prn I = 100
3. Pnr = 100 I
4. Pnr = I100

Solution: I=Pnr100 or, Pnr = 100 I

∴ Answer is Pnr = 100 I

Example 11. If the interest of 2P in t years at r2% Simple interest be I, then the interest of ? P in I2 years at r% Simple interest is

1. I2
2. 2I
3. 4I
4. I4

Solution: I=2P×t×r2×100 or, I=Prt100

∴ P×r100×t2=I2

∴ The answer is 1. I2

Example 12. The simple interest of ₹ 3000 in 2 years at the rate of 9% per annum is

1. ₹ 180
2. ₹ 270
3. ₹ 540
4. ₹ 630

Solution: 

∴ The answer is 3. ₹ 540.

Class 10 Arithmetic Chapter 1 Mcqs With Answers

Example 13. The interest of a certain amount of money in x year at x% Simple interest is ₹ x². Then the amount is

1. ₹ 100
2. ₹ 100x
3. ₹ 100x2
4. ₹ 10x

Solution: Px⋅x100=x2,P=100x2x2=100

∴ The answer is 1. ₹ 100

Example 14. If the ratio of the Principal and amount for 1 year (i.e. Principal + interest) is 20: 21, then the rate of interest p.a. is

1. 4%
2. 412%
3. 5%
4. 512%

Solution: Let Principal be ₹ 20x and amount be ₹ 21x, (x > 0)

∴ Interest = ₹ (21x – 20x) = ₹ x

∴ x=20x×1×r100, r = 5

∴ 3. The answer is 5%

WBBSE Class 10 Maths Arithmetic Chapter 2 Compound Interest And Uniform Rate Of Increase Or Decrease Multiple Choice Questions

Example 1. The interest on Rs. 700 for 2 years at 10% compound interest per annum is

1. ₹ 147
2. ₹ 126
3. ₹ 126
4. ₹ 105

Solution: Answer is 1. ₹ 147

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 2. The time in which the simple interest and compound interest will be same on a certain sum of money in an equal rate of interest is

1. 3 years
2. 2 years
3. 1 year
4. \(\frac{1}{2}\) year

Solution: Answer is 3. 1 year

Example 3. The depreciation rate of a Machine that cost Rs. 200 in 1st year is 15% and 2nd year is 10%. The cost of the machine after 2 years will be

1. Rs. 150
2. Rs. 147
3. Rs. 157
4. Rs. 153

Solution: Answer is 4. Rs. 153

Geometric Progression Mcqs With Solutions

Example 4. The present price of a machine is Rs. x and the rate of depreciation is 4y% per annum. The price of the machine will be after Z years is

1. \(x\left(1+\frac{y}{25}\right)^z\)
2. \(x\left(1-\frac{y}{25}\right)^z\)
3. \(2 x\left(1-\frac{y}{100}\right)^z\)
4. \(2 x\left(1-\frac{y}{25}\right)^{2 z}\)

Solution: Answer is 2. \(x\left(1-\frac{y}{25}\right)^z\)

Example 5. A gets Rs. 720 as an amount after 2 years from a Bank on a sum of Rs. 500. The rate of compound interest per annum is

1. 20%
2. 17\(\frac{1}{2}\)%
3. 15%
4. 10%

Solution: Answer is 3. 15%

Example 6. In case of compound interest the rate of compound interest per annum is

1. Equal
2. Unequal
3. Both equal and unequal
4. None of these

Solution: Answer is 3. Both equal and unequal

Example 7. At present the population of a village is p and if increase rate of population per year be 2r%, the population will be after n years

1. \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
2. \(2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)
3. \(\mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)
4. \(2 P\left(1-\frac{r}{100}\right)^{2 n}\)

Solution: Answer is 2. \(2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)

Class 10 Arithmetic Chapter 2 Mcqs With Answers

Example 8. The present price of a machine is ₹ 2P and if price of the machine decreased by 2r% in each year, the price of machine will be

1. \(₹ \mathrm{P}\left(1-\frac{r}{100}\right)^n\)
2. \(₹ 2 \mathrm{P}\left(1-\frac{r}{50}\right)^n\)
3. \(₹ P\left(1-\frac{r}{100}\right)^{2 n}\)
4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)

Solution: Answer is 4. \(₹ 2 \mathrm{P}\left(1-\frac{r}{100}\right)^{2 n}\)

Example 9. A person deposited ₹ 100 in a Rank and got the amount ₹ 121 for 2 years the rate of compound interest is

1. 10%
2. 20%
3. 5%
4. 10\(\frac{1}{2}\)%

Solution: Answer is 1. 10%

Class 10 Maths Arithmetic Chapter 2 MCQs

Example 10. The price of a machine decrease at a rate of x% from its previous years value. If the present price is ₹ P, the price of the machine 2 years ago was

1. \(₹ \left(\mathrm{P}-\mathrm{P} \times \frac{2 x}{100}\right)\)
2. \(₹ \frac{P}{\left(1-\frac{x}{100}\right)^2}\)
3. \(₹ \frac{P}{\left(1+\frac{x}{100}\right)^2}\)
4. \(₹ P\left(1-\frac{x}{100}\right)^2\)

Solution: \(\mathrm{P}=?\left(1-\frac{x}{100}\right)^2 \quad ?=\frac{\mathrm{P}}{\left(1-\frac{x}{100}\right)^2}\)

∴ Answer is \(₹ \frac{P}{\left(1-\frac{x}{100}\right)^2}\).

Example 11. If the compound interest in 1 year of a certain principal at a certain rate per annum be ₹ x and the simple interest for 1 year is ₹ y then

1. x > y
2. x < y
3. x = y
4. x > y

Solution: Simple interest for a certain amount for 1 year at any rate = compound interest for that amount for 1 year at the same rate of interest.

∴ x = y

∴ Answer is x = y

Example 12. If the rate of compound interest is r% per annum and the principal at the end of first year be ₹ P then the principal at the beginning, of the third year is

1. \(₹ P\left(1+\frac{r}{100}\right)^3\)
2. \(₹ \mathrm{P}\left(1+\frac{r}{100}\right)^2\)
3. \(P\left(1+\frac{r}{100}\right)^2\)
4. \(₹ P\left(1+\frac{r}{100}\right) \frac{r}{100}\)

Solution: Principal at the beginning of 2nd year = ₹ P.

∴ Principal at the beginning of 3rd year = \(₹ P\left(1+\frac{r}{100}\right)\)

∴ Answer is 3. \(₹ P\left(1+\frac{r}{100}\right)^2\)

Geometric Progression Mcqs Class 10

Example 13. In case of compound interest if the principal be ₹ P, rate of interest per annum = r, time = n years, and the final amount A, then

1. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n\)
2. \(A=P\left(1+\frac{r}{100}\right)\)
3. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n-\mathrm{P}\)
4. none of these

Solution: Answer is 1. \(\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n\)

Example 14. The fixed interval for which the interest is paid is called

1. Interest
2. Interest period
3. Amount
4. Time

Solution: Interest period

∴ Answer is 2. Interest period

WBBSE Class 10 Maths Arithmetic Chapter 3 Partnership Business Multiple Choice Questions

Example 1. The capitals of three friends in a partnership business are ₹ 200, ₹ 150, and ₹ 250 respectively. After some time the ratio of their profit share will be

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

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Solution: Ratio of profit = Ratio of the capitals

= 200: 150: 250 = 4: 4: 5

∴ The answer is 2. 4: 3: 5

Example 2. An invested Rs. 500 for 9 months and B invested Rs. 600 for 6 months. If total profit is Rs. 162, then B’s share of profit is

1. 90
2. 45
3. 36
4. 72

Solution: Ratio of equivalent capitals of A and B = (500 x 9) : (600 x 6) = 45 : 36 = 5 : 4

∴ Profit of B =

∴ Answer is 4. 72

Compound Interest Mcqs Class 10

Example 3. X invests 7 1000 for 9 months and Y invests some money for 6 months in a business. If total profit of the business is ₹ 243 and X gets ₹ 135 as profit, then Y’s capital in the business is

1. ₹ 600
2. ₹ 1200
3. ₹ 2400
4. ₹ 4800

Solution: Let Y invests ₹ P

Ratio of equivalent capitals = (1000 x 9) : (P x 6) = 1500 : P

‍∴ \(\frac{1500}{1500+P}\) x 243 = 135

or, P = 1200

∴ Answer is 2. ₹ 1200

Example 4. S and N started a business with capitals of ₹ 1500 and ₹1000. After a year then was a loss of ₹ 75. Loss of S is

1. ₹ 45
2. ₹ 30
3. ₹ 25
4. ₹ 40

Solution: Ratio of capitals = 1500 : 1000 = 3: 2

loss of S = \(\frac{3}{5}\) x ₹ 75 = ₹ 45

∴ Answer is 1. ₹ 45

Example 5. Fatima, Shreya, and Smita started a business by investing a total ₹ 6000. After a year Fatima, Shreva, and Smita get profit share of ₹ 50, ₹ 100, and ₹ 150 respectively. Smita invested in the business

1. ₹ 1000
2. ₹ 2000
3. ₹ 3000
4. ₹ 4000

Solution: Ratio of profit = 50 : 100 : 150 = 1:2:3

Smita invested = ₹ \(\left(\frac{3}{6} \times 6000\right)\)= ₹ 3000

∴ Answer is 3. ₹ 3000

Class 10 Arithmetic Chapter 3 Mcqs With Answers

Example 6. Amal and Bimal started a business. Amal invested ₹ 500 for 9 months and Bimal invested some money for 6 months. If they make a profit of ₹ 69 in a year and Bimal gets a profit of ₹ 46. The capital of Bimal is

1. ₹ 1500
2. ₹ 3000
3. ₹ 4500
4. ₹ 6000

Solution: Ratio of equivalent capitals = (500 x 9) : (x x 6) = 750 : x

∴ \(\frac{x}{x+750} \times 69=46\) = 46

or, x = 1500

∴ Answer is 1. ₹ 1500

Example 7. Pallabi invested ₹ 500 for 9 months and Raj invested ₹ 6000 for 5 months in a business. The ratio of their profit share will be

1. 3: 2
2. 5: 6
3. 6: 5
4. 9: 5

Solution: Ratio of profit = Ratio of equivalent capitals

= (500 x 9) : (600 x 5) = 45 : 30 = 3 : 2

∴ Answer is 1. 3: 2

Example 8. A invests ₹ 1800 for some time and B invests ₹ 1000 for 9 months. If profit of A and B is same, find for how many months does A invest money

1. 1 year
2. 6 months
3. 8 months
4. 5 months

Solution: Let the required time be x months.

Ratio of equivalent capitals = 1800 x : (1000 x 9) = 18x : 90 = x : 5

∴ \(\frac{x}{x+5}=\frac{5}{x+5}=x=5\)

∴ Answer is 4. 5 months

Wbbse Class 10 Maths Arithmetic Notes

Example 9. Profit of a business is Rs 1500. Capitals of first partner is ₹ 6000 and profit is ₹ 900. Capital of 2nd partner is

1. ₹ 4000
2. ₹ 6000
3. ₹ 9000
4. ₹ 15000

Solution: Ratio of capital = 6000 : x

ATP \(1500 \times \frac{6000}{6000+x}=900\)

or, x = 4000

∴ Answer is 1. ₹ 4000

Example 10. Capital of two friends is Rs 1000 and ₹ 1500 respectively. Profit of first friend is ₹ 200. Profit of 2nd friend is

1. ₹ 300
2. ₹ 400
3. ₹ 9000
4. ₹ 15000

Solution: Ratio of capitals = Ratio of profit

or, 1000 : 1500 = 200 : ?

or, ? = 300

∴ Answer is 1. ₹ 300

WBBSE Class 10 Maths Algebra Chapter 1 Quadratic Equations With One Variable Multiple Choice Questions

Example 1. The sum of two roots of the equation x² – 6x + 2 = 0

1. 2
2. -2
3. 6
4. -6

Solution: Answer is 3. 6

Example 2. If the product of two roots of the equation x² – 3x + k = 10 is -2 then the other value of k is

1. -2
2. -8
3. 8
4. 12

Solution: Answer is 3. 8

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 3. If two roots of the equation ax² + bx + c = 0 (a ≠ 0) are real and unequal then b² – 4ac will be

1. >0
2. =0
3. <0
4. None of these

Solution: Answer is 1. >0

Class 10 Maths Algebra MCQs

Example 4. If two roots of the equation ax² – bx – c = 0 (a ≠ 0) be equal then

1. c = –\(\frac{b}{2 a}\)
2. c = – \(\frac{b^2}{4 a}\)
3. c = \(\frac{b}{2 a}\)
4. c = \(\frac{b^2}{4 a}\)

Solution: Answer is 4. c = \(\frac{b^2}{4 a}\)

Example 5. If the two roots of the equation 3x² + 8x + 2 = 0 be and then the value of (\(\frac{1}{\alpha}+\frac{1}{\beta}\)) is

1. –\(\frac{3}{8}\)
2. \(\frac{2}{3}\)
3. -4
4. 4

Solution: Answer is 3. -4

Example 6. If x + 1 = 0 then x⁹ – 1 = ?

1. -2
2. 2
3. 1
4. 0

Solution: Answer is 1. -2

Example 7. kx² + 4x + 1 = 0, if roots are real and unequal of the Given quadratic equation then

1. K < 4
2. K > 4
3. K ≤ 4
4. K ≥ 4

Solution: Answer is 1. K < 4

Algebra Multiple Choice Questions Class 10

Example 8. Solution of 5x² + 2x – 7 = 0 are x = \(\frac{k \pm 12}{10}\), then k =?

1. 2
2. -2
3. \(\frac{7}{5}\)
4. 11

Solution: Answer is 2. -2

Example 9. If the roots of the equation x² + 2px + q = 0 are real and unequal then

1. p² > q
2. p² > q²
3. p² <q
4. p² < q²

Solution: Answer is 1. p² > q

Example 10. x² – ax -6 = 0 and x² + ax – 2 = 0 have a common root then a = ?

1. 0
2. 1
3. 2
4. 4

Solution: Answer is 1. 0

Example 11. No. of real roots of 3x² + 4 = 0 is

1. 0
2. 1
3. 2
4. 4

Solution: Answer is 1. 0

Wbbse Class 10 Maths Algebra Notes

Example 12. If sum of roots of x² + px +1=0, p ( > 0) is twice the difference of the roots then P =?

1. \(-\frac{1}{4}\)
2. \(\frac{3}{4}\)
3. ± \(\frac{4}{\sqrt{3}}\)
4. \(\frac{\sqrt{3}}{2}\)

Solution: Answer is 3. \(\frac{4}{\sqrt{3}}\)

WBBSE Class 10 Maths Algebra Chapter 2 Ratio And Proportion Multiple Choice Questions

Example 1. The fourth proportion of 3, 4, and 6 is

1. 8
2. 10
3. 12
4. 24

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Solution: 3 : 4 : : 6 : ? or, \(\frac{3}{4}\) = \(\frac{6}{?}\)

⇒ or, ? = \(\frac{6 \times 4}{3}\) = 8

∴ The correct answer is 1. 8

Example 2. The 3rd proportion of 8 and 12 is

1. 12
2. 16
3. 18
4. 20

Solution: 8 x third proportional = 12²

⇒ or, third proportional = \(\frac{144}{8}\) = 18

∴ The correct answer is 3. 18

Example 3. The mean proportional of 16 and 25 is

1. ± 400
2. ± 100
3. ± 20
4. ± 40

Solution: Mean proportional = ± \(\sqrt{16 \times 25}\) = ± 20

∴ The correct answer is 3. ± 20

Wbbse Class 10 Maths Algebra Notes

Example 4. a is a positive integer and a: \(\frac{27}{64}\) = \(\frac{3}{4}\): a, then the value of a is

1. \(\frac{81}{256}\)
2. 9
3. \(\frac{9}{16}\)
4. \(\frac{16}{9}\)

Solution: \(\frac{a}{\frac{27}{64}}=\frac{3}{\frac{4}{a}}\)

⇒ or, \(a^2=\frac{27}{64} \times \frac{3}{4}\)

⇒ or, a = \(\frac{9}{16}\)

∴ The correct answer is 3. \(\frac{9}{16}\)

Example 5. If 2a = 3b = 4c then a: b: c is

1. 3: 4: 6
2. 4: 3: 6
3. 3: 6: 4
4. 6: 4: 3

Solution: Let 2a = 3b = 4c = k (≠0)

∴ a : b : c = \(\frac{k}{2}: \frac{k}{3}: \frac{k}{4}\) =6:4:3

∴ The correct answer is 4. 6: 4: 3

Example 6. If 2x = 3y = 5z then x : y : z

1. 15: 10: 6
2. 6: 10: 15
3. 15: 6: 10
4. none of these

Solution: Let 2x = 3y = 5z = k (≠0)

∴ x: y: z = \(\frac{k}{2}: \frac{k}{3}: \frac{k}{4}\) = 15 : 10 : 6

∴ The correct answer is 1. 15: 10: 6

Example 7. If x + y = z and 2x – z = y then x: y: z =

1. 1: 2: 3
2. 2: 1: 3
3. 3: 1: 2
4. none of these

Solution: x = z – y = \(\frac{y+z}{2}\)

⇒ or, 2z- 2y = y + z or, z = 3y

⇒ Now, x = z- y = 3y-y = 2y

∴ x: y: z = 2y: y: 3y = 2: 1: 3

∴ The correct answer is 2. 2: 1: 3

Class 10 Maths Algebra Chapter 2 MCQs

Example 8. If \(\frac{m}{n}\) = \(\frac{2}{3}\) then \(\frac{n+m}{n-m}\) is

1. 2
2. 4
3. 5
4. 10

Solution: \(\frac{n}{m}=\frac{3}{2} \quad \text { or, } \quad \frac{n+m}{n-m}=\frac{3+2}{3-2}\) [componendo & dividendo]

⇒ or, \(\frac{n+m}{n-m}=5\)

∴ The correct answer is 3. 5

Example 9. If (a + b) : √ab = 4 : 1 then a : b =

1. (2+√3):(2-√3)
2. (2-√3) : (2+√3)
3. 1:1
4. None of these

Solution: \(\frac{a+b}{\sqrt{a b}}=\frac{4}{1}\)

⇒ or, \(\quad \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{4+2}{4-2}\) [Componendo & dividendo]

⇒ or, \(\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=3\)

⇒ or, \(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\sqrt{3}\), Again by componendo and dividendo,

⇒ We get, \(\frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\),

⇒ \(\frac{a}{b}=\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)^2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\)

∴ The correct answer is 1.

Example 10. If \(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{2 a-3 b+c}{p}\) = \(\frac{2 a-3 b+c}{p}\) then P =

1. 0
2. 1
3. 2
4. 3

Solution: Let \(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}\) = k (≠ 0)

∴ a = 3k, b = 4k, c = 7k.

∴ \(\frac{2 \cdot 3 k-3 \cdot 4 k+7 k}{p}=k\)

⇒ or, \(\frac{k}{p}\)

∴ p = 1

∴ The correct answer is 2. 1

WBBSE Class 10 Maths Algebra Chapter 3 Quadratic Surds Multiple Choice Questions

Example 1. If x = 2 + √3 then the value of x + \(\frac{1}{x}\) is

1. 2
2. 2√3
3. 4
4. 2-√3

Solution: Answer is 3. 4

Wbbse Class 10 Maths Chapter 3 MCQs

Example 2. If peq = √13, p-q = √5 then the value of pq is

1. 2
2. 18
3. 9
4. 8

Solution: Answer is 1. 2

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Example 3. If a + b = √5, a-b = √3 then a² + b² = ?

1. 8
2. 4
3. 2
4. 1

Solution: Answer is 2. 18

Class 10 Algebra Chapter 3 Mcqs With Answers

Example 4. If we subtract √5 from √125 the value is

1. 80
2. 120
3. 100
4. none of these

Solution: Answer is 1. 80

Example 5. The product of (5 – √3) (√3-1) (5 + √3) (√3 +1) is

1. 22
2. 44
3. 2
4. 11

Solution: Answer is 2. 44

Example 6. a = √6 + √5, ab = 1 then a² + b² = ?

1. 24
2. 26
3. 20
4. 22

Solution: Answer is 4. 22

Example 7. Greatest of 0, √5 + √3 and √6 + √2 is

1. 0
2. √5 + √3
3. √6+√2
4. None of these

Solution: Answer is 2. √5 + √3

Linear Equations Mcqs With Solutions

Example 8. √50 – 3√2-√8 =?

1. 1
2. √2
3. √5
4. 0

Solution: Answer is 4.0

Example 9. √5-√3=a,√5+√3=?

1. \(\frac{2}{a}\)
2. \(\frac{a}{2}\)
3. 0
4. none of these

Solution: Answer is 1. \(\frac{2}{a}\)

Example 10. If (√7- √3) ÷ √5 = \(\frac{1}{5}\) \((\sqrt{35+x})\) = then x = ?

1. √15
2. 15
3. -√15
4. -15

Solution: Answer is 3. -√15

WBBSE Class 10 Maths Algebra Chapter 4 Variation Multiple Choice Questions

Example 1. x ∝ \(\frac{1}{y}\) then

1. x = \(\frac{1}{y}\)
2. y = \(\frac{1}{x}\)
3. xy = 1
4. xy = non-zero constant

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

Solution: x ∝ \(\frac{1}{y}\) ⇒ xy=k

⇒ (k is a non-zero variation constant)

∴ The correct answer is 4. xy = non-zero constant

Example 2. If x ∝ y then

1. x²  ∝ y³
2. x³ ∝ y²
3. x  ∝ y³
4. x² ∝ y²

Solution: x = ky (k is a non-zero variation constant)

⇒ or, \(\frac{x^2}{y^2}=k^2\)

∴ x² ∝ y²

∴ The correct answer is 4. x² ∝ y²

Class 10 Maths Algebra Chapter 4 MCQs

Example 3. If x ∝ y and y = 8 when x = 2 if y = 16 then the value of x is

1. 2
2. 4
3. 6
4. 18

Solution: x = ky (k is a non-zero variation constant)

⇒ or, 2 = k x 8 or, k = \(\frac{1}{4}\)

∴ x = \(\frac{1}{4}\)y = \(\frac{1}{4}\) x 16 = 4

∴ The correct answer is 2. 4

Example 4. If x ∝ y² and y = 4 when x = 8 ; if x = 32 then the value of y is

1. 4
2. 8
3. 16
4. 32

Solution: x = ky² (k is a non-zero variation constant)

⇒ or, 8 = k(4)²

⇒ or, k = \(\frac{1}{2}\)

⇒ Now, x = \(\frac{1}{2}\)y²

⇒ 32 = \(\frac{1}{2}\)y²

⇒ or, y = 8

∴ The correct answer is 2. 8

Example 5. If y – z ∝ \(\frac{1}{x}\) , z – x ∝ \(\frac{1}{y}\), x – y ∝ \(\frac{1}{z}\) sum of three variation constant is

1. 0
2. 1
3. -1
4. 2

Solution: y – z = \(\frac{k_1}{x}\), z – x = \(\frac{k_2}{y}\) , x – y = \(\frac{k_3}{z}\) (k₁, k₂, k₃ are non zero variation constant)

⇒ Now, k₁ + k₂ + k₃ = xy – xz + yz – xy + xz – yz = 0

∴ The correct answer is 1. 0

Factorization MCQs With Solutions

Example 6. If x ∝ \(\frac{1}{y}\) & y ∝ \(\frac{1}{z}\) then

1. x ∝ z
2. xy ∝ z
3. x ∝ yz
4. x ∝ \(\frac{1}{z}\)

Solution: x = \(\frac{k_1}{y}\) , y = \(\frac{k_2}{z}\) (k₁, k₂ non zero variation constant)

⇒  x = \(\frac{k_1}{y}\) = \(\frac{k_1}{\frac{k_2}{z}}\) = \(\frac{k_1}{k_2}\)

∴ x ∝ z

∴ The correct answer is 1. x ∝ z

Example 7. y ∝ x² and y = 9 when x = 9 then the value of x if y = 4

1. ± 3
2. ± 4
3. ± 6
4. ± 8

Solution: y = kx² (k is a non zero variation constant)

⇒ 9 = k (9)² or, k = \(\frac{1}{9}\)

∴ y = \(\frac{1}{9}\)x²

⇒  x² = 36

⇒  or, x = ± 6

∴ The correct answer is 3. ± 6

Example 8. Corresponding value of x and y are x = 12, y = 16 ; x = 3, y = 4 then relation between x and y is

1. x  ∝ \(\frac{1}{y}\)
2. x³ ∝ y²
3. x³ ∝ y²
4. x² ∝ y²

Solution: \(\frac{x}{y}=\frac{12}{16}=\frac{3}{4}, \quad \frac{x}{y}=\frac{3}{4}\)

∴ \(\frac{x}{y}\) = constant, x ∝ y

∴ The correct answer is 2. x² ∝ y²

Class 10 Algebra Chapter 4 MCQs With Answers

Example 9. If x ∝ y³ and z² ∝ x then

1. y ∝ z³
2. y ∝ z²
3. y³ ∝ z²
4. y³ ∝ z

Solution: x ∝ y³ or, x = k₁y³, z² = k₂x (k₁, k₂ non zero variation constant)

∴ xz² = k₁k₂ y³x

⇒  or, \(y^3=\frac{1}{k_1 k_2} z^2\)

⇒  or, y³ ∝ z²

∴ The correct answer is 3. y³ ∝ z²

Example 10. If a² ∝ bc, b² ∝ d, c² ∝ b, d² ∝ ac and abed = k then value of k is

1. ∝ ab
2. ∝ bc
3. ∝ cd
4. ∝ da

Solution: \(a^2=k_1 b c, b^2=k_2 d, c^2=k_3 b, d^2=k_4 a c\) (k₁, k₂, k₃, k₄ non zero variation constant)

a²b²c²d² = k₁, k₂, k₃, k₄ (abcd) bc, k ∝ bc or, (abcd) ∝ bc

∴ The correct answer is 2. ∝ bc

WBBSE Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions

Example 1. If 3x = cosec α and \(\frac{3}{x}\) = cot α, then the value of 3\(\left(x^2-\frac{1}{x^2}\right)\) is

1. \(\frac{1}{27}\)
2. \(\frac{1}{81}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{9}\)

Solution: \((3 x)^2-\left(\frac{3}{x}\right)={cosec}^2 \alpha-\cot ^2 \alpha\)

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

⇒ \(9\left(x^2-\frac{1}{x^2}\right)=1 \quad \Rightarrow 3\left(x^2-\frac{1}{x^2}\right)=\frac{1}{3}\)

∴ The Correct Answer is 3. \(\frac{1}{3}\)

Example 2. If 2x = sec A and \(\frac{2}{x}\) = tan A, then the value of 2\(\left(x^2-\frac{1}{x^2}\right)\) is

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{8}\)
4. \(\frac{1}{16}\)

Solution: \((2 x)^2-\left(\frac{2}{x}\right)^2=\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\)

⇒ \(4\left(x^2-\frac{1}{x^2}\right)=1\)

⇒ \(2\left(x^2-\frac{1}{x^2}\right)=\frac{1}{2}\)

∴ The correct answer is 1. \(\frac{1}{2}\)

Class 10 Maths Trigonometry Chapter 2 MCQs

Example 3. If tan α+ cot α = 2 then the value of (tan¹³ α + cot¹³ α) is

1. 1
2. 0
3. 2
4. None of these

Solution: \(\tan \alpha+\cot \alpha=2\)

⇒ \(\tan \dot{\alpha}+\frac{1}{\tan \alpha}=2\)

⇒ \(\frac{\tan ^2 \alpha+1}{\tan \alpha}=2\)

⇒ \(\tan ^2 \alpha+1=2 \tan \alpha \Rightarrow \tan ^2 \alpha-2 \tan \alpha+1=0\)

⇒ \((\tan \alpha-1)^2=0\)

⇒ \(\tan \alpha-1=0 \quad \Rightarrow \quad \tan \alpha=1\)

∴ \(\cot \alpha=\frac{1}{\tan \alpha}=1\)

= \((1)^{13}+(1)^{13}=1+1=2\)

∴ The correct answer is 3. 2

Example 4. If sin θ- cos θ = 0 [0° ≤ θ ≤ 90°] and sec θ + cosec θ = x, then the value of x is

1. 1
2. 2
3. 2
4. 2√2

Solution: sin θ – cos θ = 0

⇒ sin θ = cos θ

⇒ \(\frac{\sin \theta}{\cos \theta}=1\)

⇒ tan θ = 1 = tan 45° [0° ≤ θ ≤ 90°]

⇒ θ = 45°

x = sec θ + cosec θ = sec 45° + cosec 45°

= √2 + √2 = 2√2

∴ The correct answer is 4. 2√2

Heights And Distances Mcqs With Solutions Class 10

Example 5. If 2 cos 3θ = 1, then the value of θ is

1. 10°
2. 15°
3. 20°
4. 30°

Solution: 2 cos 3θ= 1

⇒ cos 3θ = \(\frac{1}{2}\)

⇒ cos 3θ= cos 60°

⇒ 3θ= 60° ⇒ θ = 20°

∴ The correct answer is 3. 20°

Example 6. If sin θ + cosec θ =2 then the value of (sin θ – cosec θ) is

1. 2
2. 1
3. 0
4. \(\frac{\sqrt{3}}{2}\)

Solution: sin θ + cosec θ = 2

(sin θ – cosec θ)² = (sin θ+ cosec θ)² – 4 sin θ.cosec θ

= (2)² – 4 x 1 = 0

⇒ sin θ – cosec θ = 0

∴ The correct answer is 3. 0

Example 7. If tan θ = \(\frac{a}{b}\) then the value of sin θ is

1. \(\frac{a}{\sqrt{a^2-b^2}}\)
2. \(\frac{b}{\sqrt{a^2+b^2}}\)
3. \(\frac{a+b}{\sqrt{a^2+b^2}}\)
4. \(\frac{a}{\sqrt{a^2+b^2}}\)

Solution: tan θ = \(\frac{a}{b}\)

⇒  cot θ = \(\frac{b}{a}\)

cosec θ = \(\sqrt{1+\cot ^2 \theta}=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{a^2+b^2}}{a}\)

sin θ = \(\frac{a}{\sqrt{a^2+b^2}}\)

∴ The correct answer is 4. \(\frac{a}{\sqrt{a^2+b^2}}\)

Class 10 Trigonometry Chapter 2 Mcqs With Answers

Example 8. If sec θ – cosec θ = 0 then the value of tan θ is

1. 0
2. 1
3. -1
4. 2

Solution: sec θ – cosec θ = 0

⇒ sec θ = cosec θ

⇒ \(\frac{1}{\cos \theta}=\frac{1}{\sin \theta}\)

⇒ cos θ = sin θ

tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sin \theta}\) = 1

∴ The correct answer is 2. 1