## Railway RRB Group D 2018 Mathematics Quiz

Railway RRB Group D Mathematics Quiz 2018. Welcome to the www.letsstudytogether.co online Railway RRB Group D section. If you are preparing Railway, RRB ALP and RRB Group D Exams 2018, you will come across a section on “General Mathematics”. Here we are providing you with “Important RRB Group D 2018 Mathematics Quiz ”   based on the latest pattern of your daily practice.

This “Railway RRB Group D 2018 Mathematics Quiz “ is also important for other Railway exams such as RRB ALP & Technicians Exam 2018.

## RRB Group D 2018 Mathematics Quiz | Set-9

1.A salesman is allowed a discount 5.5% on the total sales made by him plus a bonus of 0.5% on sales over Rs. 10,000. If his total earnings were Rs. 1,990, then his total sales (in Rs.) were:

A. Rs. 34,000

B. Rs. 45,000

C. Rs. 22,000

D. Rs. 32,000

A. Rs. 34,000

Let the salesman’s total sales be Rs. (10,000 + x), then,

According to the question,

(10,000 × 5.5% + x × 6%) = 1990

⇒ 5000 × (11/100) + x × (6/100) = 1990

⇒ 5000 × 11 + 6 × x = 1,99,000

⇒ 6x = 199000 – 55000

⇒ 6x = 144000

⇒ x = 144000/6 = 24,000

∴ Total sales = Rs. (10,000 + 24,000) = Rs. 34,000

Hence, the required total sales were Rs. 34,000.

2. The unit digit in the product 17 × 23 × 51 × 93 is

A. 1

B. 3

C. 5

D. 7

B. 3

Unit’s digit = 7 × 3 × 1 × 3 = 3

3.The least number that must be added to 8961 to make it exactly divisible by 84 is

A. 27

B. 57

C. 141

D. 107

A. 27

When 8961 is divided by 84, the reminder is 57.

∴ The least number that must be added to 8961 to make it exactly divisible by 84

= (84 – 57) = 27.

4.Amar can complete a work in 30 days and Raman can complete the same work in 15 days. If both of them work together, then in 4 days what percent of the total work will be completed?

A. 15

B. 37

C. 40

D. 45

C. 40

Work done by Amar in 1 day is 1/30th of total work

Work done by Raman in 1 day is 1/15th of the total work

According to the problem statement, work done in 4 days can be given as

⇒ work done = 4 × (1/30 + 1/15)

Hence, 4/10 of the total work will be completed

⇒ percentage of total work = 4 × 100/10 = 40%

∴ the 40% of total work will be completed in 4 days

5.The sum of three digits of a number is 15. If 198 is added to the number, the digits in the units and hundreds place are interchanged. Also, the digit at tens place is mean of the other two digits. What is the number?

A. 654

B. 456

C. 753

D. 357

B. 456

Let the digits be X, Y and Z.

Therefore, the number is (100X + 10Y + Z).

From 1st condition,

(100X + 10Y + Z) + 198 = (100Z + 10Y + X)

⇒ 99Z – 99X = 198

⇒ Z – X = 2 — (i)

From 2nd condition,

Y = (X + Z)/2 — (ii)

Also, X + Y + Z = 15

∴ X + X/2 + Z + Z/2 = 15

⇒ X + Z = 10 —- (iii)

⇒ Y = 5

From (i) and (iii), we get X = 4 and Z = 6

∴The number is 456.

6.Find the value of question mark (?):

[23 ÷ (1/2)] – ?% of √5625 = 34 + [4 × (3/8) – (2/3) ÷ (4/27)]

A. 20

B. 18

C. 25

D. 15

A. 20

Follow BODMAS rule to solve this question, as per the order given below,

Step – 1 – Parts of an equation enclosed in ‘Brackets’ must be solved first,

Step – 2 – Any mathematical ‘Of’ or ‘Exponent’ must be solved next,

Step – 3 – Next, the parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated,

Step – 4 – Last but not least, the parts of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated

⇒ [23 ÷ (1/2)] – ?% of √5625 = 34 + [4 × (3/8) – (2/3) ÷ (4/27)]

⇒ 46 – ?% of 75 = 34 + [4 × (3/8) – 9/2]

⇒ 46 – ?% of 75 = 34 + [(3/2) – (9/2)]

⇒ 46 – ?% of 75 = 34 – 3

⇒ 46 – (?/100) × 75 = 34 – 3

⇒ 46 – 3 × ?/4 = 31

⇒ ? = (46 – 31) × 4/3

⇒ ? = 15 × 4/3

∴ ? = 20

7.If 57 = 47, 76 = 55 and 86 = 62, what is 99 = ?

A. 95

B. 96

C. 99

D. 94

C. 99

→ 57 = (5 × 7) + (7 + 5) = 47

→ 76 = (7 × 6) + (7 + 6) = 55

→ 86 = (8 × 6) + (8 + 6) = 62

Similarly,

⇒ 99 = (9 × 9) + (9 + 9) = 99

8.Value of the expression 4 of (5/4) + [-6 – (-12) of 2] is

A. 20

B. 21

C. 23

D. 30

C. 23

Follow BODMAS rule to solve this question, as per the order given below,

Step – 1 – Parts of an equation enclosed in ‘Brackets’ must be solved first,

Step – 2 – Any mathematical ‘Of’ or ‘Exponent’ must be solved next,

Step – 3 – Next, the parts of the equation that contain ‘Division’ and ‘Multiplication’ are calculated,

Step – 4 – Last but not least, the parts of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.

Using BODMAS rule

4 of (5/4) + [-6 – (-12) of 2] = 5 + [-6 – (-24)] = 5 + 18 = 23

9. Rajneesh covers 1/3rd of his journey at a speed of 25 km/hr and half of the rest at a speed of 45 km/hr and the rest with 10 km/hr, then find the average speed?

A. 15.5 kmph

B. 16.5 kmph

C. 10.5 kmph

D. 18.5 kmph

D. 18.5 kmph

We know that average speed, when distance of three intervals is same is 3uvw/(uv + vw + wu)

From the given data, we get

⇒ Average speed = (3 × 25 × 45 × 10)/(25 × 45 + 45 × 10 + 25 × 10)

= (30 × 25 × 45)/1825

= 18.5 kmph

∴ Average speed for the whole journey is 18.5 kmph

10.For an article the profit is 190% of the cost price. If the cost price increase by 10% but the selling price remains same, then profit is what percentage of selling price (approximately)?

A. 54

B. 62

C. 70

D. 163

B. 62

Let its cost price be Rs. Y

⇒ selling price = y × (1 + 190/100)

⇒ selling price = 2.9y

Cost after 10% increase in CP

⇒ Cost = y (1 + 10/100)

⇒ Cost = 1.1y

Profit is given as

⇒ (2.9y – 1.1y) × 100/2.9y

⇒ 180/2.9

⇒ 62 (approx.)

∴ profit is 62% of selling price