Mathematics Question For RRB NTPC & RRC Group D Exams 2019 | Set- 2

 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.

B. 3

Unit’s digit = 7 × 3 × 1 × 3 = 3

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.

 C. 40

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

Work done by Raman in 1 day is 1/15^th 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

 B. 456

Let the digits be X, Y and Z.

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

From 1^st condition,

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

⇒ 99Z – 99X = 198

⇒ Z – X = 2 — (i)

From 2^nd 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.

 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

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

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

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

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