__Numerical Ability Practice Questions for IBPS RRB (Set – 4)__**1. A rise of 25% in the price of oranges compels a person to buy 50 oranges less for Rs. 500. Find the increased price of oranges per dozen.**

A) 24

B) 26

C) 28

D) 30

**2. Two sides of a rectangle are measured. The measure of one side is too big by 4% of its true length and that of the other is too short by 5% of its true length. Find the error percent in the measure obtained for the area of rectangle?**

A) 6/5%

B) 5/6%

C) 4/3%

D) 3/4%

E) 1%

**3. A batsman's average score for a number of innings was 21.75 per inning. In the next three innings he scored 28, 34, 37 runs and his average for all the innings was raised by 1.125. How many innings in all did he play?**

A) 24

B) 30

C) 36

D) 40

E) 42

**4. By selling 12 articles for Rs. 60 a man loses an amount equal to selling price of 3 articles. Find the loss percent.**

A) 15%

B) 18%

C) 20%

D) 23%

E) 25%

**5. A tradesman allows a discount of 15% on the marked price. How much above the cost price must he mark his goods to make a profit of 19%?**

A) 35%

B) 40%

C) 45%

D) 50%

E) 55%

**6. 1, 2, 8, 48,?**

A) 384

B) 364

C) 394

D) 374

E) 344

**7. 12, 13, 17, 26, 42, ?**

A) 57

B) 67

C) 63

D) 61

E) 59

**8. 1, 4, 18, 44, 83, ?**

A) 132

B) 134

C) 136

D) 138

E) 140

**9. 8, 4, 4, 6, 12, ?**

A) 24

B) 26

C) 28

D) 30

E) 32

**10. 157, 150, 136, 115, 87,?**

A) 48

B) 62

C) 58

D) 64

E) 52

**Solutions:**

**1. D)**Increase in amount due to price rise = 25% of Rs.500 = Rs.125

Increase price of 50 oranges = Rs.125

Increase price per dozen = 125/50 × 12 = Rs.30

**2. A)**Net percentage change in area = P1 + P2 + [(P1 × P2)/100] = 4 – 5 + [(4 × - 5)/100] = -1 – 1/5 = – 6/5%

**3. B)**Let he played x innings.

x × 21.75+ 28 + 34 +37 = (x + 3) × (1.125 + 21.75)

=> x = 30.375/1.125 = 27

Therefore no. of innings = 27 + 3 = 30

**4. C)**loss% = [3/(12 + 3)] × 100 = 20%

**5. B)**Let the Cost price be Rs.100

Therefore SP = Rs.119

Therefore MP = 119/85 × 100 = Rs.140

Therefore Required % = [(140 – 100)/100] × 100 = 40%

**6. A)**1×2….2×4….8×6….48×8

**7. B)**1….4…..9….16…..25

**8. C)**Double difference

3…..14…..26…..39

11…….12….13…….14

**9. D)**x0.5….x1…..x1.5…..x2…..x2.5

**10. E)**-7…..-14…..-21….-28