SSC MTS Numeric Aptitude Mock Test in Hindi Series 2, SSC Online Test

सुत्रानुसार
N = शंकुओं की संख्‍या
N × 1/3π × 3 × 3 × 4 = 4/3 × π × 6 × 6 × 6
N = 24

प्रश्‍नानुसार
दुकानदार 20% कम वजन का प्रयोग करता है ।
तो वह काम में लेता है = 1000 × 80/100 = 800 gm
लाभ = 200/800×100 = 25%
अभीष्‍ट लाभ = 10 + 25 + 25×10/100 = 37.5%

अभीष्‍ट मूलधन = 
= 1×100×365/5
= Rs. 7300

Given, the ratio of milk to water in the first container is 2 : 1
Let, Amount of milk in the first container = 2x
And, amount of water in the first container = x
The ratio of milk to water in the second container is 9 : 5
Let, Amount of milk in the second container = 9y
Amount of water in the second container = 5y
∵ Capacity of both containers are equal
∴ 2x + x = 9y + 5y
⇒ 3x = 14y
⇒ x/y = 14/3 ……. (A.
If two containers are mixed,
Amount of milk in the mixture= 2x + 9y
Amount of water in the mixture = x + 5y
∴ Required ratio
= 2x + 9y/x+5y

Using A

∴Decrease per cent = 10×10/100 = 1%

Arbind’s income = Rs. 100
Expenditure = Rs. 75
Saving = Rs. 25
New income = Rs. 120
Expenditure
= 75 + 7.5 = Rs. 82.5
Savings = 120 – 82.5 = Rs. 37.5
Required percentage
= 37.5-25/25 × 100 = 50%

Let present age of daughter be x years, then
her mother’s age = (46 – x) years
After 4 years,
Daughter’s age = (x + 4) years and
Mother’s age = (46 – x + 4) years = (50 – x)years
According to question,
50-x = 2×(x+4)
⇒ 50-x = 2x + 8
⇒ 3x = 42
⇒x = 42/3 = 14 years
Thus, daughter’s present age = 14 years
Hence, option A is correct.

20% of A = 30% of B = 1/6 of C
⇒ A = 5k, B = 10/3k, C = 6k
∴  A : B : C = 5k : 10k/3 : 6k = 15 : 10 : 18

Given,
9 pipes can fill up the swimming pool in 20 minutes.
∴ Part of swimming pool fill by 1 pipe in 1 minutes = 1/(9 × 20) = 1/180
Let, ‘x’ pipe can fill up the swimming pool in 18 minutes.
∴ (1/180) × x × 18 = 1
⇒ x/10 = 1
⇒ x = 10

Difference = Pr² /(100)² 
= P×16/1000
P = 1000/16 = ₹ 625

=>2250/9x – 225/x = 25
=>250/x – 225/x = 25
x = Re 1 per kg
Hence Option B is correct

Let the marked price be ₹ x.
∴ x × 75/100 = 200×135/100
x = 200×135/75 = ₹ 360

Here speeds of two trains = 16 miles/hour and 21 miles/hour respectively.
So I can say in an hour second train runs 5 miles more than the first train.
Now at the time of their meeting, the second train has travelled 60 miles more then the first. So I can say trains meet after 12 hours(60/5)
Hence distance between A & B = (16+21)(12) = 444 miles
Hence Option A is correct

If B completes a work in x days.
A will complete the same in 2x / 3 days.
∴  1/x + 3/2x = 1/10
⇒ 2+3 / 2x = 1/10 ⇒ 2x = 50
⇒ x = 25 days

If the length of journey be x km, then

LCM of 16,20,24 =240
Now 240=2×2×2×2×3×5

Hence the required least number which is a perfect square and divisible by 16,20,24 =2×2×2×2×3×5×3×5=3600

The mean value of 20 observations was found to be 75
Sum of these 20 values = 75 × 20 = 1500
Mistakenly ‘97’ was read as ‘79’.
Hence difference = 97 – 79 = 18
Hence actual sum = 1500+18 = 1518
Therefore actual average = 1518 / 20 = 75.9

∴ Diameter of cylinder = Diameter of cone