# Rukmini Railway Group-D Mathematics Vol-1

**Rukmini Railway Group-D Mathematics Vol-1**

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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

The volume of 1st part, an ∝ 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 grams instead of 100 grams 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.

**Rukmini Railway Group-D Mathematics Vol-1**

**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 question are solved with an explanation. The questions are based on 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.**

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