Quantitative Aptitude - Arithmetic Ability Questions

As now, the time 4 : 48 pm after 45 min it will be

4:48 + 45 min = (4 hrs : 48 + 12 min) pm + 33 min = 5 hrs : 33 min pm.

Hence, it is 5:33 pm.

P=2(l+b)

l=4b

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

⇒14+112=13

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.

Ratio of girls and boys = 100% : 75% = 4:3

Difference is 1 unit = 35

So, 3 units = 35 x 3 = 105

If 20% absent, that’s mean 80% present.

So, present boys = 80% of 105 = 80×105100 = 84

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

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

Given day is Wednesday.

 Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, ∠ACB = 30°and∠ADB =45°

ABAC=tan30°=13=>AC=AB*3=1003m

 ABAD=tan45°=1=>AD=AB=100m

CD=(AC+AD)=1003+100m=1003+1=100*2.73=273m

0.75 = 75/100 = 3/4