OSSSC Math Test – 01

Given:

A and B  take 5 days

B and C take 7 days

A and C take 4 days

Solution:

LCM(5, 7, 4) = 140 let total work be 140 unit

The efficiency of A, B, C are

⇒ 2(A + B + C) = 28 + 20 + 35 = 83

⇒ A + B + C = 83/2 = 41.5

⇒ A = (A + B + C) - (B + C) = 41.5 - 20 = 21.5

By the same process

⇒ B = 41.5 - 35 = 6.5

⇒ C = 41.5 - 28 = 13.5

⇒ Time(A) = 140/21.5 = 6.51

⇒ Time(B) = 140/6.5 = 21.54

⇒ Time(C) = 140/13.5 = 10.37

A took the least time to complete the work. So, A is the fastest.

∴ A finishes fastest in 6.51 days.

Given:

Amount(A) = Rs. 44,100

Rate = 10% per annum

Time = 2 years

Calculation:

A = 44,100

Rate = 10% pa and time = 1 year

If interest is compounded half-yearly, then

Rate = 10/2 = 5% and t = 2

As we know,

44,100 = P(1 + r/100)²

⇒ 44,100 = P × (1 + 5/100)²

⇒ P = 44,100 × 20/21 × 20/21

⇒ A = 40,000

Given : 

The leak emptied the cistern in 8 hours.

The pipe with the leak emptied the cistern = 12 hours.

Calculation : 

To find the cistern capacity, take LCM of 8 and 12 i.e. 24.

Efficiency of outlet Pipe = 24/8 =  -3 units per hour

Efficiency of outlet + Inlet Pipe = 24/12 = - 2 units per hour

Efficiency of Inlet Pipe = -2 -(- 3) = 1 unit per hour

∴Time taken by the pipe to fill the cistern

⇒ 24/1 = 24 hours

The correct answer is option 2 i.e. 24 hours

Given :

Pen is sold at a gain of 6% instead of at a loss of 6%

The seller gets Rs. 6 more

Formula Used :

Profit = Selling Price - Cost Price / 100

Calculation:

The cost price of an article is “x”

Then sold at 6% gain ⇒ x + 0.06x = 1.06x

Sold at 6% loss ⇒ x – 0.06x = 0.94x

& their difference is Rs. 6

⇒ 1.06x – 0.94x = 6

⇒ x = Rs. 50

∴ The correct answer is Rs. 50

Given:

HCF of two numbers = 4

LCM of two numbers = 80

One of the numbers is 4 more than the other

Formula used:

Product of two numbers = HCF × LCM

Calculation:

Let the smaller number be 4k

Then, the larger number = (4k + 4)

Now, the Product of two numbers = HCF × LCM

⇒ 4k(4k + 4) = 4 × 80

⇒ k(k + 1) = 20

k = 4 and it satisfy the equation.

Smaller number = 4k

⇒ 4(4) ⇒ 16

∴ The answer is 16.

Given:

8% more is gained by selling an article for ₹300 than by selling it for ₹280.

Formula Used:

Profit = SP - CP

Loss = CP - SP

SP = CP × {(100 + Profit%)/100}

where,

SP = Selling Price,

CP = Cost Price

Calculation:

According to the question,

Let the CP be 100x

The increase = 100x × 8% = 8x

SP increase = 300 - 280 = 20

⇒ 8x → 20

x → 20/8 = 2.5

100x → 2.5 × 100 = 250

The answer is 250.

Given:

The average of two numbers A and B is 20, that of B and C is 19 and C and A is 21.

Concept Used:

Average = Sum of Observation/Tota number of Observation

Calculation:

Average of two number A and B is 20

Sum of two number = 20 × 2 = 40      ---- 1

Average of two number B and C is 19

Sum of two number = 19 × 2 = 38      ---- 2

Average of two number C and A is 21

Sum of two number = 21 × 2 = 42      ---- 3

Add equation 1, 2 and 3

⇒ 2(A + B + C) = 40 + 38 + 42

⇒ A + B + C = 60      ---- 4

According to the question,

The Value of A

Subtract equation 2 from equation 4

⇒ A = 60 - 38

⇒ 22

∴ The value of A is 22.

Given

Total time is taken whole journey = 10 hrs.

The speed at whole journey = 48 km per hr.

Formula Used

Distance = Speed × Times

Distance = 48 × 10 = 480 kms

Now, this distance of 480 km is to be covered in 8 hours. Hence, the required speed

⇒ 480/8 = 60 km pe hr

Increased speed = 60 - 48 = 12 km/h

∴ The speed of the car must be increased by 12 km/h

Given∶

Percentage of water in fresh fruit = 72%

Percentage of water in dry fruit = 20%

Total Weight of fresh fruit = x = 100 kg

Calculation∶

Let the weight of the total weight of dry fruit be T.

Percentage of fruit pulp in fresh fruit = 28%

Weight of fruit pulp in fresh fruit = 28% of x = 28/100 × 100 = 28 kg

Percentage of fruit pulp in dry fruit = 80%

Weight of fruit pulp in dry fruit = 80% of T

We know that the weight of fruit pulp will be constant in both fresh fruit and dry fruit.

∴ 28 kg = 80% of T

⇒ T = 28 × 100/80 = 35 kg

Let the total number of votes polled be n

Given, one candidate got 40% of votes,

⇒ Number of votes received by loser = 40% of n = 40n/100

⇒ Votes % got by winner = 100 – 40 = 60%

So, Number of votes by winner = 60n/100

According to the question,

⇒ (60n/100) – (40n/100) = 298

⇒ 20n/100 = 298

⇒ n/5 = 298

⇒ n = 298 × 5 = 1490

∴ Total number of votes polled is 1490.

Formula used:

HCF of different fractions = HCF of numerators / LCM of denominators

Calculations:

HCF = HCF of (3,4,7)/LCM of (4,5,60)

⇒ 1/60

Calculation:

⇒ 1800 ÷ 10 × [45 ÷ (17 - 2)] of 2 + [- 2 (1 + 2)]

⇒ 1800 ÷ 10 × [45 ÷ 15] of 2 + [- 6]

⇒ 1800 ÷ 10 × 3 of 2 + [- 6]

⇒ 180 × 6 - 6

⇒ 1080 - 6

⇒ 1074

∴ The correct answer is 1074.