Quantitative Aptitude - Arithmetic Ability Questions

Number = ( 779 x a) + 47, where "a" is the quotient

           = (19 x 41 x a) + (19 x 2) + 9

           = 19 x (41a + 2) + 9

           = 19 x (New quotient) + 9

  Required remainder = 9

Speed of the boat downstream  s=a/t= 60/3 = 20 kmph

Speed of the boat upstream s= d/t = 30/3= 10 kmph

Therefore,  The speed of the stream = speed of downstream - speed of upstream2=5 kmph

Given speed = 63 km/hr = 63×518=352 m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (170 + x)mts is 30 sec.

We know speed = distancetimem/s

⇒352=170+x30

 -->  x = 355 mts.

side of first square = 40/4 =10cm

side of second square = 32/4 = 8cm

area of third squre = 10² - 8² = 36 sq cm

side of third square = 36 = 6 cm

required perimeter = 6*4 = 24cm

Deepa invested Rs.50,000 for 12 months, Rs.(50000 + 20000) for 12 months and Rs.(50000 + 20000 + 20000) for 12 months.
i.e, she invested Rs.50,000 for 12 months, Rs.70000 for 12 months and Rs.90000 for 12 months.
Sushma invested Rs. 70000 for 2 years; i.e, Rs.70000 for 24 months
And, Avanthi invested Rs.70000 for 1 year; i.e, Rs. 70000 for 12 months.

Their investing ratio:
Deepa : Sushma : Avanthi = (50,000 x 12 + 70000 x 12 + 90000 x 12):(70000 x 24):(70000 x 12)
= (25,20,000):(16,80,000):(8,40,000) = 252:168:84 = 3:2:1

Total profit for 3 years = Rs.3,00,000
Therefore, Sushma's share = Rs.(3,00,000 x 2 /(3+2+1)) = Rs.(3,00,000 x 2/6) = Rs.1,00,000.

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

The 5 consecutive odd numbers whose average is k are  (k-4), (k-2), k, (k+2), (k+4)

Again the average of (k-4), (k-2), k, (k+2), (k+4), (k+6), (k+8) is (k+2)

Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways