Top 10 Maths solutions of questions for competitive exams that you must solve

In this article, you are getting top 10 maths solutions of questions that will help you in many exams like SSC Cgl, SSC CHSL, MTS, Railways, Bank, all state exams, and many other competitive exams in India. These questions will help you for quick revision in the last days of the exam.

Here, you will also get a short and simple solution that will help you for saving time in exams. Know about some important solutions of questions that can be used in your various competitive Exams, which are as follows:-

1. A shopkeeper allows 3 successive discounts of 50%, 40% and 20% respectively. Find equivalent discount?

(a) 66% (b) 70%

(c) 60% (d) 76%

Solution. Here, we will solve this problem by the ratio method. The equivalent ratio will be equal to the composition of many ratios.

Ratio form of given percentage-

50% = -1/2, 40% = -4/10 =-2/5, 20% = -1/5

Ratios- 2 : (2-1) = 2:1 

               5 : (5-2) = 5:3 

               5 : (5-1) = 5:4 

         __________________

              2×5×5 : 1×3×4

                     50 : 12 

% discount = (50-12)×100/50 = 76% 

Ans. D

2. If sin 7x = cos 11x, then the value of tan 9x + cot 9x is

(a) 1 (b) 2

(c) 3 (d) 4

Solution. If Sin 7x = Cos 11x

                     Sin 7x = Sin(90° -11x)

                         7x = 90° - 11x

                         18x = 90°

                            X = 5°

Put x = 5° in tan9x + cot9x 

                   = tan45° + cot45° = 1+1 = 2

Ans. B

3. P, Q, R are employed to do work for Rs. 5750. P and Q together finished 19/23 of work, and Q and R together finished 8/23 of work. The wage of Q, in rupees, is-

(a) 2850 (b) 3750

(c) 2750 (d) 1000

Solution. We know Work done by P + Q + R = 1

 Work done by Q = (P+ Q) + (Q + R) - (P+Q+R)

                               = 19/23 + 8/23 – 1

                               = 4/23

The wage of Q = 4/23 * 5750 = 4 × 250

                           = Rs. 1000 

Ans. D

4. If 738A6A is divisible by 11, then the value of A is-

(a) 6 (b) 3

(c) 9 (d) 1

 Solution. The divisibility rule of 11 is If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

(A + A + 3) - (6 + 8 + 7) = 2A -18

2A - 7 - 11 

Since -11 is divisible by 11 so remove this 

So 2A - 7 is divisible by 11

That is 2A - 7 = 11

               A = 18/2 = 9

Ans. C

5. A shopkeeper marks up his goods 35% above the CP and gives 23% discount to the customer. at the time of buying he uses 1120 gm instead of 1kg and At the time of selling the goods he gives 880 gm weight instead of 1kg .find his profit %?

(a) 35.66% (b) 23.76%

(c) 33.3% (d) 32.3%

Solution. We will solve this question by the ratio method. We know that the equivalent ratio of mark up, discount, the dishonest weight will be equal to the actual profit/loss percentage.

Mark up = 35% = +35/100 = 100 : 135 = 20 : 27

Discount = 23% = -23/100 = 100 : 77

At the time of buying and selling goods = less : more (always)

At the time of buying = 1000 : 1120 = 25 : 28

At the time of selling = 880 : 1000 = 22 : 25

Equivalent ratio is 20×100×25×22 : 27×77×28×25

           20×100×22 : 27×77×28

               5×100×2 : 27×7×7

                     1000 : 1323

 Profit % = (1323-1000)×100/1000 

                = 32.3%

Ans. D

6. If x² + x + 1 = 0, then x²⁰¹⁸ + x²⁰¹⁹ equals which of the following

(a) x (b) –x 

(c) x –1 (d) x + 1

Solution. X² + x + 1 = 0

        x + 1/x = -1

         x³ + x = -x² = -(-x -1)

         x³ + x = x + 1

         x³ = 1

So x²⁰¹⁸ + x²⁰¹⁹ = x² + 1 = -x - 1 + 1 = -x

Ans. B

7. The ratio of the length of two trains is 4 : 3, and the ratio of their speed is 6: 5. The ratio of time taken by them to cross a pole is?

(a) 5 : 6 (b) 11 : 8

(c) 10 : 9 (d) 27 : 16

Solution. Since we know that 

Time = distance/speed

So the ratio of time taken by them to cross a pole is = 4/6: 3/5

    = 20 : 18 = 10 : 9

Ans. C

8. A spherical metal ball of 6 cm radius is melted and recast into three spherical balls. The radii of two of these balls are 3 cm and 4 cm. What is the radius of the third ball?

(a) 4.5 cm (b) 5 cm

(c) 6 cm (d) 7 cm

Solution. We know that

Volume of big spherical metal ball = 3 recasted small spherical balls

4/3 * π 6³ = 4/3 * π 3³ + 4/3 * π 4³ + 4/3 * π r³

6³ = 3³ + 4³ + r³

216 = 27 + 64 + r³

r³ = 216 - 91

r³ = 125

r = 5 cm

Ans. B

9. The difference between the compound interest and simple interest on rupees x at 13% per annum for 2 years is 81.12 rupees. What is the value of x?

(a) 4800 (b) 4600 

(c) 5400 (d) 4750           

Solution. Principal × (Rate/100)² = difference between the compound and simple interest for 2 years

Here principal is x

So x × (13/100)² = 81.12

       x × 169/10000 = 81.12

       x = 81.12×10000/169

       x = 4800

Ans. A

10. If a student scores 43% marks, he fails by 72 marks, but when he scores 57% marks, he is passed by 40 marks. Find passing %?

a)53% b)55% 

c)52% d)50%   

Solution. We will solve this question by the unit method.

43% marks + 72 marks = 57% marks - 40 marks

(57 - 43)% marks = (72 + 40) marks

14% = 112 

1% = 8

9% = 72 

So passing marks = 43% + 9% = 52% 

Ans. C