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

1. Algebra.
2. Arithmetic.
3. Calculus.
4. Geometry.
5. Probability and Statistics.
6. Number System.
7. Set Theory.
8. Trigonometry.
9. Number System
10. H.C.F & L.C.M
11. Decimal Fraction
12. Simplification
13. Square Root & Cube Root
14. Average
15. Problem on Numbers
16. Problem on Ages
17. Surds & Indices
18. Percentage
19. Profit & Loss
20. Ratio & Proportion
21. Partnership
22. Compound Proportion
23. Time & Work
24. Pipes & Cistern
25. Time And Distance
26. Problem on train
27. Boats & Streams
28. Allegation & Mixture
29. Simple Interest
30. Compound Interest
31. Area
32. Volume of Solids
33. Races
34. Calendar
35. Clock
36. Stock & Shares
37. True Discount
38. Algebra
39. Banker’s Discount
40. Linear Education in Two Variables
41. Quadratic Equation
42. Trigonometry
43. Line & Angles
44. Triangles
45. Quadrilaterals
46. Circle
47. Polygons
48.  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) x2 + 5x – 20,

(b) x2 – 4x – 20,

(c) x2 – x – 20,

(d) x2 + x – 20.
Solution:

(x – 4)(x + 5).

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

= x2 + 5x – 4x – 20.

= x2 + x – 20.

Answer: (d)

Q7. Factor: 5x2 – 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:

5x2 – 15x – 20.

= 5(x2 – 3x – 4).

= 5(x2 – 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:(4x2 – 2x) – (-5x2 – 8x).
Solution:

(4x2 – 2x) – (-5x2 – 8x)

= 4x2 – 2x + 5x2 + 8x.

= 4x2 + 5x2 – 2x + 8x.

= 9x2 + 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 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?
Solution:

33/4 ÷ 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: 16w3 – u4w3

Solution:

16w3 – u4w3.

= w3(16 – u4).

= w3(42 – ((u2)2).

= w3(4 + u2)(4 – u2).

= w3(4 + u2)(22 – u2).

= w3(4 + u2)(2 + u)(2 – u).

Answer: w3(4 + u2)(2 + u)(2 – u).

Q14. Factor: 3x4y3 – 48y3.
Solution:

3x4y3– 48y3.

= 3y3(x4 – 16).

= 3y3[(x2)2 – 42].

= 3y3(x2 + 4)(x2 – 4).

= 3y3(x2 + 4)(x2 – 22).

= 3y3(x2 + 4)(x + 2)(x -2).

Answer: 3y3(x2 + 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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