Quantitative Aptitude - Arithmetic Ability Questions

No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360

Total time taken = k/40 + 2k/20 hours

= 5k/40 = k/8 hours

Average speed = 3k/(k/8) = 24 kmph.

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton 

i.e. 3 engines of latter type consumes 1 ton for one hour

Hence 9 engines consumes 3 tons for one hour

For 15 hours it is 15 x 3 = 45 metric tonnes.

Given C.I = 5984, R = 20% , T = 2yrs

 C.I = P1+201002-1

5984 = 

=> P = (5995x25)/11

P = Rs. 13625

Now S.I = PTR/100

SI = (13625 x 8 x 6)/100 = Rs. 6540

Speed = 78 x 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (900 + x)/60 = 65/3
x = 400 m.

Manager's monthly salary

= Rs. (1900 x 25 - 1500 x 24) = Rs. 11,500

Shopkeeper sells at CP using weight of 900 gm instead of1 kg

So, 1000gm –900 gm = 100gm

⇒100/900 X 100

⇒100/9 %

Let, Distance between A and B = d

Distance travelled by P while it meets Q = d + 11

Distance travelled by Q while it meets P = d – 11

Distance travelled by Q while it meets R = d + 9

Distance travelled by R while it meets Q = d – 9

Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11

The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9

But given Ratio of speeds of P & R => P : R = 3 : 2

 SPSR = SPSQxSQSR = d+11d+9d-11d-9

=> d+11d+9d-11d-9 = 3/2

=>  d = 1, 99

=> d = 99 satisfies.

Therefore, Distance between A and B = 99