Test: Arithmetical Reasoning - 2 - UPSC MCQ

Let the number of boys be x. Then, (3/4)x = 18 or x = 18 x (4/3) = 24.
If total number of students is y, then (2/3) y = 24 or y = 24 x (3/2) = 36.
Therefore Number of girls in the class = (36 - 24) = 12.

Let son's age be x years. Then, father's age = (3x) years.
Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.
So, 3x - 5 = 4 (x - 5)  3x - 5 = 4x - 20  x = 15.

Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.

L.C.M. of 6, 5, 7, 10 and 12 is 420.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

Originally, let number of women = x. Then, number of men = 2x.
So, in city Y, we have: (2x - 10) = (x + 5) or x - 15.
Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 45.

Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A+C = 2E ...(iii)

B + D = A+C = 2E ...(iv)

A+B + C + D + E = 150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

Since each pole at the corner of the plot is common to its two sides, so we have :

Total number of poles needed = 27 x 4 - 4 = 108 - 4 = 104.

We have : A = 3B ...(i) and

C - 4 = 2 (A - 4) ...(ii)

Also, A + 4 = 31 or A= 31-4 = 27.

Putting A = 27 in (i), we get: B = 9.

Putting A = 27 in (ii), we get C = 50.

Let Varan's age today = x years.

Then, Varan's age after 1 year = (x + 1) years.

Therefore x + 1 = 2 (x - 12)  x + 1 = 2x - 24  x = 25.

There were all sparrows but six' means that six birds were not sparrows but only pigeons and ducks.

Similarly, number of sparrows + number of ducks = 6 and number of sparrows + number of pigeons = 6.

This is possible when there are 3 sparrows, 3 pigeons and 3 ducks i.e. 9 birds in all.

Clearly, the smallest such number is 3.

Three ducks can be arranged as shown above to satisfy all the three given conditions.