Quantitative Aptitude - Arithmetic Ability Questions

Given that the diameters of the three pipes are 2 cm, 3 cm and 4 cm

From the given data,

Amount of water from three pipes is 4 units, 9 units and 16 units.

Let the capacity of cistern be 'p' units.

∴ p/58 = 16

⇒ p = 928 units.

In 1 minute, quantity to be filled by 3 pipes = 29 units

∴ Total time required = 928/29 = 32 minutes.

Total savings = 200 + 200 + 200 + 240 + 280 + ... to n months = 11040

= 120 + 80 + 160 + 40 + 200 + 240 + ....n terms
= 40 + 80 + 120 + 160 + 200 + 240 + ...n terms
= n/2[2a + (n-1)d] + 80 + 40 = 11040
= n/2[240 + (n-1)40] = 10920
= n[6+n-1] = 546
= n[n+5] = 546
n = 21

30 + (.0825 x 30) + (15% x 30)

Now

=> 0.0825 x 30 = 2.475 -----(1)

=> 15% of 30 = 450/100 = 4.5 -----(2) 

Hence, from (1) & (2)

30 + (.0825 x 30) + (15% x 30)

= 30 + 2.475 + 4.5

30 + (.0825 x 30) + (15% x 30) = 36.975.

As given in the question, Marked price is 25% more than the Cost price.

=> C.P of the article = 34 x 400 = 300

Now, 

Let the original S.P of the article be Rs. P

Now the new S.P = P + 16.666 +13.333300 x P

=> S.P = 7P6

According to the question,

7P6 - 300 = 2P - 300

=> 5P = 1800

=> P = Rs. 360

Hence, the increased S.P = 360 x 7/6 = Rs. 420. 

πr2 =26.64

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days

167 days = (23 weeks + 6 days) =6 odd days.

Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.

Given day is Wednesday.