# Crack IBPS PO : data interpretation : Aptitude day 56

# Crack IBPS PO : data interpretation : Aptitude day 56

**D.1-5) Chart 1 shows the distribution of twelve million tonnes of courier transport through different modes over a specific period of time. Chart 2 shows the distribution of the cost of transporting this courier. The total cost was Rs.30 million.**

Q.1) What is the cost of transporting courier by rail per ton(in Rs)?

a) 2.5

b) 3.33

c) 6.4

d) 8

e) None of these

Q.2) If the cost per ton of transport by ship, airfreight and road are represented by P, Q and R respectively, which of the following is true?

a) R > Q > P

b) P > R > Q

c) P > Q > R

d) R > P > Q

e) None of these

Q.3) Which is the most effective way of transportation?

a) Road

b) Ship

c) Pipeline

d) can’t be determined

e) None of these

Q.4) If the pipeline cost increases by 30% , by what percentage would revenue have to be increased so as to have same amount of profit?

a) 11.5

b) 12.8

c) 13.75

d) 14.6

e) None of these

Q.5) If for some reason ship stop sailing, by what percentage the airfreight have to go up to reach the previous level of volume transported (approximately)?

a) 75%

b) 81%

c) 85%

d) 90%

e) None of these

**D.6-10) ****What approximate value will come in place of question mark (?) in the given question? (You are not expected to calculate the exact value).**

Q.6) 105.27 % of 1200.11 + 11.80% of 23600.85 = 21.99% of (?) + 1420.99

a) 5000

b) 2400

c) 3100

d) 12140

e) 9600

Q.7) 0.98% of 7824 + 4842 ÷ 119.46 – (?) = 78

a) 30

b) 60

c) 40

d) 50

e) 70

Q.8) (42.1992 –18.1642 )^{(2 )}-(?) = 168.62 – 138.99

a) 4004

b) 1200

c) 172

d) 867

e) 546

Q.9) (?)^{3}+ √8– 340 = (√8×√8^{)(1/2)}+ (9)^{(1/2)}

a) 7

b) 19

c) 18

d) 9

e) None of these

Q.10) (15 × 0.40)^{4}÷ (1080 ÷ 30)^{4} × (27 × 8)^{4} = (3 × 2)^{?}×6^{5}

a) 8

b) 3

c) 12

d) 16

e) None of these

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## 2 comments

Day 13 reasoning??

posted.sorry for delay.

post link : http://bankersdaily.in/target-sbi-po-reasoning-day-13/

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