Quantitative Aptitude - Arithmetic Ability Questions

Let the age of sushma be x and
the age of her son is y
Then five years before x-5=5(y-5) ...(1)
Five years hence x+5 = 3(y+5)-8 .....(2)

By soving (1) & (2), we get
5y - 15 = 3y + 7
y = 11 => x = 35

Therefore, the age of Sushma = 35 and her son = 11.

                      Article I                   Article II
                         -20                         25%
                                        5%
                         20%                       25%
                     

                          4              :              5

Now, Selling of first article = (4-20%) = 0.8

Selling of second article = (5+25%) =  6.25

Therefore the ratio of selling price = 8 : 62.5

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

So, required probability = 11!×3!13! = 39916800×66227020800 = 24625

Given the angles of a quadrilateral are in the ratio of 2:4:7:5
Let the angles of a quadrilateral are 2x, 4x, 7x, 5x
But we know that sum of the angles = 360 degrees.
=> 2x + 4x + 7x + 5x = 360
=> x = 20
Therfore, the smallest angle of the quadrilateral = 2x = 2x20 = 40 degrees.
One of the angle of the triangle = 2 x 40 = 80 degrees
The other angle is 180 - (40 + 80) = 60 degrees.

Hence the second largest angle of the triangle is 60 degrees.

As given Kapil's 8th birthday was celebrated 2 years ago,

Kapil’s present age = 8 + 2 = 10 years.

Father's age is twice of Kapil’s age after 10 years.
=> F + 10 = 2 (K + 10) = 2 (10+ 10)
F = 30years,

Avanthi's age =1/6(F) = 30/6 = 5 years.

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

Time from 12 p.m on monday to 2 p.m on the following monday = 7 days 2 hours = 170 hours

Therefore, The watch gains 2+445 min. or  34/5 min. in 170 hrs.

Now, 34/5 min. are gained in 170 hrs.

Therefore, 2 min are gained in 170×534×2 hrs = 50 hrs

Therefore, Watch is correct 2 days 2 hrs. after 12 p.m on monday i.e it will be correct at  2 p.m on wednesday.