# RS Aggarwal Maths Solution Notes in hindi

**RS Aggarwal Maths Solution Notes in Hindi **

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Today we are sharing an important note pdf** RS Aggarwal Maths Solution Notes in Hindi** – SSC NOTES like SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many other examsArihant English Book PDF 2022 Latest Grammar Composition is very important for any competitive exam and this **RS Aggarwal Maths Solution Notes in Hindi SSC****, CGL, CPO & Other Competitive Exams **for SSC, BANK, AND RAILWAYS is very useful for it. this FREE PDF will be very helpful for your examination.

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** MATH TOPICS**

**Algebra.****Arithmetic.****Calculus.****Geometry.****Probability and Statistics.****Number System.****Set Theory.****Trigonometry.****Number System****H.C.F & L.C.M****Decimal Fraction****Simplification****Square Root & Cube Root****Average****Problem on Numbers****Problem on Ages****Surds & Indices****Percentage****Profit & Loss****Ratio & Proportion****Partnership****Compound Proportion****Time & Work****Pipes & Cistern****Time And Distance****Problem on train****Boats & Streams****Allegation & Mixture****Simple Interest****Compound Interest****Area****Volume of Solids****Races****Calendar****Clock****Stock & Shares****True Discount****Algebra****Banker’s Discount****Linear Education in Two Variables****Quadratic Equation****Trigonometry****Line & Angles****Triangles****Quadrilaterals****Circle****Polygons****Number Series**

** MATH QUESTION AND ANSWER**

**Q1 A right circular cone is cut by 3 planes parallel to its base. The planes cut the altitude of the cone in four equal parts. Find out the ratio of volume of each part.**

A

1 : 7 : 19 : 37

B

1 : 8 : 27 : 64

C

1 : 9 : 16 : 25

D

1 : 2 : 3 : 4

» Explain it

Correct Option: A

Volume of cone, V ∝r3 ∝h3

Volume of 1st part, a ∝ h3

Volume of 2nd part, b ∝ 8h3– h3 = 7h3

Volume of 3rd part, c∝ 27h3– 8h3 =19h3

Volume of 4th part, d ∝ 64h3– 27h3 = 37h3

Therefore, required ratio = h3 ∶ 7h3 ∶ 19h3 ∶ 37h3 = 1 ∶ 7 ∶ 19 ∶ 37

Hence, option A is correct.

**Q2 -A merchant uses a weight of 125 gram instead of 100 gram while buying an article. He used 80 grams instead of 100 grams while selling. He marked up the price by 20% and then offers 20% discount. Find the overall profit or loss percentage.**

A

20%

B

30%

C

40%

D

50%

» Explain it

Correct Option: D

Effective ratio of SP to CP

= 125 × 100 × 120 × 80 = 3

100 80 100 100 2

Therefore Profit Percentage = 1 ×100 = 50 %

2

Thus , D is the correct answer.

3 -If x =

√5 + 1 ,

√5 – 1

then the value of 5×2 – 5x – 1 will be

A

0

B

3

C

4

D

5

» Explain it

Correct Option: C

According to the given question

we have,

x =

√5 + 1 =

√5 + 1

√5 – 1

√5 – 1

Now, Numerator and denominator multiplied by (√5 + 1)

x =

(√5 + 1) × (√5 + 1)

(√5 – 1) (√5 + 1) =

(√5 + 1)2

5 – 1 = √5 + 1

2

Now we also have, 5×2 – 5x – 1

put value of x in above equation

⇒ 5 ( √5 + 1 ) 2 – 5 ( √5 + 1 ) – 1

2 2

⇒ 5 (6 + 2√5) – 5√5 – 5 – 1

4 2

⇒ 5 (3 + √5) – 5√5 – 5 – 2

2 2

⇒ 15 + 5√5 – 5√5 – 7 = 8 = 4

2 2

Hence, option C is correct.

**Q4 The ratio of the work done by 50 women to the work done by 25 men, in the same time is 4 : 3. If 18 women and 12 men can finish a piece of work in 5 days, then how many women can finish the same work in 20/3 days?**

A

18

B

27

C

33

D

30

» Explain it

Correct Option: B

Given that,

The ratio of the work done by 50 women and 25 men is

1 : 1

3 4

The ratio of the work done by one man and one woman

= 1 : 1

150 100

Let the time taken by one woman and one man to complete the work be 150x and 100x respectively.

18 × 5 + 12 × 5 = 1

150x 100x

x = 6

**Q5 Time taken by one woman and one man to complete the work be 180 days and 120 days respectively.**

The number of women worked for 20/3 days to complete the work

= 3 × 180 = 27

20

Hence, option B is correct.

**Q6 Two varieties of sugar are mixed in the ratio 3 : 2 and sold for ₹80 per kg to make a profit of 25%. If the cost of the variety of sugar whose quantity is more is ₹40 per kg, what is the cost of the other variety of sugar?**

A

Rs. 50

B

Rs. 48

C

Rs. 75

D

Rs. 100

» Explain it

Correct Option: D

Sale price of the mixture = Rs.80

Cost price of mixture = 80 × 100 = Rs. 64

125

Rs. 40 Rs. x

\ /

Rs. 64

/ \

x – 64 24

⇒ x – 64 = 3 ⇒ x = Rs. 100

24 2

Hence, option D is correct.

In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students can understand each and every step.

**1. Which is greater than 4?**

(a) 5,

(b) -5,

(c) -1/2,

(d) -25.

Solution: 5 greater than 4.

**Answer: (a)**

**2. Which is the smallest?**

(a) -1,

(b) -1/2,

(c) 0,

(d) 3.

Solution:

The smallest number is -1.

**Answer: (a)**

**Q 3. Combine terms: 12a + 26b -4b – 16a.**

(a) 4a + 22b,

(b) -28a + 30b,

(c) -4a + 22b,

(d) 28a + 30b.

Solution:

12a + 26b -4b – 16a.

= 12a – 16a + 26b – 4b.

= -4a + 22b.

**Answer: (c)**

**Q4. Simplify: (4 – 5) – (13 – 18 + 2).**

(a) -1,

(b) –2,

(c) 1,

(d) 2.

Solution:

(4 – 5) – (13 – 18 + 2).

= -1-(13+2-18).

= -1-(15-18).

= -1-(-3).

= -1+3.

= 2.

**Answer: (d)**

**Q5. What is |-26|?**

(a) -26,

(b) 26,

(c) 0,

(d) 1

Solution:

|-26|

= 26.

**Answer: (b)**

**Q6. Multiply: (x – 4)(x + 5)**

(a) x^{2} + 5x – 20,

(b) x^{2} – 4x – 20,

(c) x^{2} – x – 20,

(d) x^{2} + x – 20.

Solution:

(x – 4)(x + 5).

= x(x + 5) -4(x + 5).

= x^{2} + 5x – 4x – 20.

= x^{2} + x – 20.

**Answer: (d)**

**Q7. Factor: 5x ^{2} – 15x – 20.**

(a) 5(x-4)(x+1),

(b) -2(x-4)(x+5),

(c) -5(x+4)(x-1),

(d) 5(x+4)(x+1).

Solution:

5x^{2} – 15x – 20.

= 5(x^{2} – 3x – 4).

= 5(x^{2} – 4x + x – 4).

= 5{x(x – 4) +1(x – 4)}.

= 5(x-4)(x+1).

**Answer: (a).**

**Q8. Factor: 3y(x – 3) -2(x – 3).**

(a) (x – 3)(x – 3),

(b) (x – 3)^{2},

(c) (x – 3)(3y – 2),

(d) 3y(x – 3).

**Solution:**

3y(x – 3) -2(x – 3).

= (x – 3)(3y – 2).

**Answer: (c).**

**Q9. Solve for x: 2x – y = (3/4)x + 6.**

(a) (y + 6)/5,

(b) 4(y + 6)/5,

(c) (y + 6),

(d) 4(y – 6)/5.

Solution:

2x – y = (3/4)x + 6.

or, 2x – (3/4)x = y + 6.

or, (8x -3x)/4 = y + 6.

or, 5x/4 = y + 6.

or, 5x = 4(y + 6).

or, 5x = 4y + 24.

or, x = (4y + 24)/5.

Therefore, x = 4(y + 6)/5.

**Answer: (b).**

**Q10. Simplify:(4x ^{2} – 2x) – (-5x^{2} – 8x).**

Solution:

(4x^{2} – 2x) – (-5x^{2} – 8x)

= 4x^{2} – 2x + 5x^{2} + 8x.

= 4x^{2} + 5x^{2} – 2x + 8x.

= 9x^{2} + 6x.

= 3x(3x + 2).

**Answer: 3x(3x + 2)**

**Q11. Find the value of 3 + 2 • (8 – 3)**

(a) 25,

(b) 13,

(c) 17,

(d) 24,

(e) 15.

Solution:

3 + 2 • (8 – 3)

= 3 + 2 (5)

= 3 + 2 × 5

= 3 + 10

= 13

**Answer: (d)**

**Q12. Rice weighing 3 ^{3}/_{4} pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?**

Solution:

3^{3}/_{4} ÷ 4 pounds.

= (4 × 3 + 3)/4 ÷ 4 pounds.

= 15/4 ÷ 4 pounds.

= 15/4 × 1/4 pounds.

= 15/16 pounds.

Now we know that, 1 pound = 16 ounces.

Therefore, 15/16 pounds = 15/16 × 16 ounces.

= 15 ounces.

**Answer: 15 ounces.**

**Q13. Factor: 16w ^{3} – u^{4}w^{3}**

Solution:

16w^{3} – u^{4}w^{3}.

= w^{3}(16 – u^{4}).

= w^{3}(4^{2} – ((u^{2})^{2}).

= w^{3}(4 + u^{2})(4 – u^{2}).

= w^{3}(4 + u^{2})(2^{2} – u^{2}).

= w^{3}(4 + u^{2})(2 + u)(2 – u).

**Answer: w ^{3}(4 + u^{2})(2 + u)(2 – u).**

**Q14. Factor: 3x ^{4}y^{3} – 48y^{3}.**

Solution:

3x^{4}y^{3}– 48y^{3}.

= 3y^{3}(x^{4} – 16).

= 3y^{3}[(x^{2})^{2} – 4^{2}].

= 3y^{3}(x^{2} + 4)(x^{2} – 4).

= 3y^{3}(x^{2} + 4)(x^{2} – 2^{2}).

= 3y^{3}(x^{2} + 4)(x + 2)(x -2).

**Answer: 3y ^{3}(x^{2} + 4)(x + 2)(x -2)**

**Q15. What is the radius of a circle that has a circumference of 3.14 meters?
Solution:**

Circumference of a circle = 2πr.

Given, circumference = 3.14 meters.

Therefore,

2πr = Circumference of a circle

or, 2πr = 3.14.

or, 2 × 3.14r = 3.14,[Putting the value of pi (π) = 3.14].

or, 6.28r = 3.14.

or, r = 3.14/6.28.

or, r = 0.5.

**Answer: 0.5 meter.**

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