OSSSC Math Test – 02

Given:

A trader marks his goods at 55% above the cost price.

He allows a discount of 20%.

Concept used:

SP = CP + Profit

Profit = CP × (Profit percentage / 100)

Calculation:

Let the cost price be 100

Then the marked price will be 155

After giving a 20% Discount SP will be

⇒155 - (155 × 20/100)

= 124

The profit percentage will be 124 - 100 = 24%

Calculation:

Let, Arun's present age = 4x & Dipak's present age = 3x

Given that in 6 years, Arun's age will be 26

Arun's present age = 26 - 6 = 20 years

4x = 20 or x = 5

Substitute x = 5 into Dipak's present age:

∴ Dipak's present age = 3x = 3 × 5 = 15 years

Given

Average of 9 numbers = 15

Average of first five numbers = 13

Average of last five numbers = 17

Concept:

Average = Sum of numbers/Number of numbers.

Calculation:

Sum of 9 numbers = 9 ×  15 = 135

⇒ Sum of first five numbers = 5 ×  13 = 65

⇒ Sum of last five numbers = 5 ×  17 = 85

⇒ Now, the sum of the first five numbers and the sum of the last five numbers include the fifth number twice.

⇒ So, the fifth number = (Sum of first five numbers + Sum of last

five numbers) - Sum of all 9 numbers = (65 + 85) - 135 = 15.

Therefore, the fifth number is 15.

Given that

A + B + C = 5625    -----(1)

According to the question

A = (B + C)/2

⇒ B + C = 2 A

Adding A on both sides

⇒ A + B + C = 3A = 5625     [from (1)]

⇒ 3A = 5625

⇒ A = Rs. 1875

Again B = ( A + C )/4

⇒ A + C = 4B

⇒ A + B + C = 5B = 5625     [from (1)]

⇒ 5B = 5625

⇒ B = Rs. 1125

⇒ A - B = 1875 - 1125 = 750

∴ A's share is more than that of B by Rs. 750.

Given data:

Principal (P) = ₹ 8,300

Rate of interest (r) = 10%

Time (t) = 2 years

Concept:

Difference between compound interest and simple interest.

Solution:

Simple interest (SI) = (P × r × t)/100

⇒ ₹ 8,300 × 10% × 2 = ₹ 1660

Compound Interest (CI) = P × [(1 + r/100)^t - 1]

⇒ ₹ 8,300 × [(1 + 10/100)² - 1] = ₹ 1743

Difference = CI - SI

⇒ ₹ 1743 - ₹ 1660 = ₹ 83

Hence, the difference between CI and SI is ₹ 83.

Given:

A and B together can complete a work in 8 days.

A alone can complete it in 16 days.

B is working at 50 percent efficiency.

Concept:

Work = Rate x Time

Solution:

⇒ If A alone can complete the work in 16 days, A's work rate is 1/16.

⇒ If A and B together can complete the work in 8 days, their combined work rate is 1/8.

⇒ B's work rate is therefore (1/8 - 1/16) = 1/16.

⇒ If B is working at 50 percent efficiency, B's current work rate is 1/16 ×  0.5 = 1/32.

⇒ Therefore, A and B's combined work rate is 1/16 + 1/32 = 3/32.

⇒ Hence, A and B together will complete the work in 1 / (3/32) days.

Therefore, A and B together will complete the work in 32/3 days.

olution:

⇒ {8.4 ÷ 12 × [(27 - 6)2 ÷ 3] (7 - 5)}

⇒ {0.7 × [(21)2 ÷ 3] (2)}

⇒ {0.7 × [441 ÷ 3] (2)}

⇒ {0.7 × [147] (2)}

⇒ {0.7 × 294}

⇒ 205.8.

Given data:

Current population = 24000

Expected decrease = 4%

Concept or formula:

New population = Current population - (Expected decrease % of current population)

Calculation:

Calculate the new population

⇒ New population = 24000 - (4% of 24000) = 24000 - 960 = 23040

Hence, the population of the village next year will be 23040.

Given : 

Cost of bicycle = ₹1,200

Repairing cost = ₹1,800

Selling price = ₹2,400

Concept : 

Loss % = ((Cost - Selling Price) / Cost) × 100

Calculation : 

Total cost = 1200 + 1800 = ₹3000

Loss = Total cost - Selling price = 3000 - 2400 = ₹600

Loss % = (600 / 3000) × 100 = 20%

Therefore, the loss percentage is 20%.

Given:

Upstream speed = 8 km/h

Downstream speed = 12 km/h

Concept:

Speed in still water = (Downstream speed + Upstream speed) / 2

Speed of water = (Downstream speed - Upstream speed) / 2

Calculation:

Speed in still water = (12 + 8) / 2 = 10 km/h

Speed of water = (12 - 8) / 2 = 2 km/h

Therefore, the man's speed in still water is 10 km/h and the speed of water is 2 km/h.

Given:

Speed of the boat in still water: 10 km/h

Speed of the stream: 2 km/h

Distance to travel downstream: 72 km

Concept:

The effective speed of the boat downstream is the sum of the boat's speed in still water and the speed of the stream.

Calculation:

⇒ Speed of the boat downstream = 10 km/h + 2 km/h = 12 km/h

⇒ Time = Distance/Speed

⇒ Time = 72 km/12 km/h = 6 hours.

Therefore, the boat will take 6 hours to travel 72 km downstream.

Given : 

P and Q together can complete the work in 20 days.

Q and R together can complete it in 30 days.

P is twice as good a workman as R.

Concept used : 

Work done = Efficiency × Time, where Efficiency is the ability to do work.

Solution : 

Total work = LCM(20, 30) = 60 units

Efficiency of P and Q = 60/20 = 3      ...........(i)

Efficiency of Q and R = 60/30 = 2      ...........(ii)

Acc to given condition

Efficiency of P : Efficiency of R = 2 : 1

Adding equation (i) and (ii), we get

Efficiency of (P + 2Q + R) = 5

⇒ 2 + 2Q + 1 =

⇒ 2Q = 2

⇒ Efficiency of Q = 1

Days taken by Q to complete the work alone = 60/1 = 60 days

Given:

L₁ = 550 m, S₁ = 68 km/h

L₂ = 650 m, S₂ = 40 km/h

Calculation:

⇒ S₁' = 68 × (1000/3600) = 18.89 m/s

⇒ S₂' = 40 × (1000/3600) = 11.11 m/s

Calculate the relative speed (S) and total distance (D):

⇒ S = S₁' + S₂' = 18.89 m/s + 11.11 m/s = 30 m/s

⇒ D = L₁ + L₂ = 550 m + 650 m = 1200 m

Calculate the time (T) using T = D/S:

⇒ T = D / S = 1200 m / 30 m/s = 40 seconds

Hence, the trains will cross each other in 40 seconds

Given data:

Three fractions: 2/5, 4/15, 6/25

Concept or formula:

LCM of fractions = LCM of numerators/HCF of denominators

Calculation:

LCM of numerators (2, 4, 6) = 12

HCF of denominators (5, 15, 25) = 5

Therefore, LCM of the fractions = 12/5 = 12/5

Hence, the LCM of the fractions 2/5, 4/15, and 6/25 is 12/5.