Test: Quantitative Reasoning- 2 - SSC CGL MCQ

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.

Let total number of members be 100,

Then, number of members owning only 2 cars = 20.

Number of members owning 3 cars = 40% of 80 = 32.

Number of members owning only 1 car = 100 - (20 + 32) = 48.

Thus, 48% of the total members own one car each.

When Rahul was born, his brother's age = 6 years; his father's age = (6 + 32) years = 38 years, his mother's age = (38 - 3) years = 35 years; his sister's age = (35 - 25) years = 10 years.

Let number of horses = number of men = x.

Then, number of legs = 4x + 2 x (x/2) = 5x.

So, 5X = 70 or x = 14.

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.

Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.

Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.

Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

Let number of girls = x and number of boys = 3x.

Then, 3x + x = 4x = total number of students.

Thus, to find exact value of x, the total number of students must be divisible by 4.

'All but nine died' means 'All except nine died' i.e. 9 sheep remained alive.

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then, .

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 6R = 108 R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

Total runs scored = (36 x 5) = 180.

Let the runs scored by E be x.

Then, runs scored by D = x + 5; runs scored by A = x + 8;

runs scored by B = x + x + 5 = 2x + 5;

runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.

So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.

Therefore 3x + 120 =180 3X = 60 x = 20.