# Basic Numeracy - Online Test

Q1. A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
Explaination / Solution:

 Speed= 60 x 5 m/sec = 50 m/sec. 18 3

Length of the train = (Speed x Time). Length of the train = 50 x 9 m = 150 m. 3

Q2. A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Explaination / Solution:

 Speed of the train relative to man = 125 m/sec 10

 = 25 m/sec. 2

 = 25 x 18 km/hr 2 5

= 45 km/hr.

Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 x = 50 km/hr.

Q3. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
Explaination / Solution:

Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ACB = 30° and ADB = 45°.

 AB = tan 30° = 1 AC = AB x √3 = 100√3 m. AC √3

 AB = tan 45° = 1 AD = AB = 100 m. AD CD = (AC + AD) = (100√3 + 100) m = 100(√3 + 1) = (100 x 2.73) m = 273 m.

Q4. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
Explaination / Solution:

Let AB be the wall and BC be the ladder. Then, ACB = 60º and AC = 4.6 m.

 AC = cos 60º = 1 BC 2 BC = 2 x AC = (2 x 4.6) m = 9.2 m.

Q5. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Explaination / Solution:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698.

Q6. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
Explaination / Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

 Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508 100 100 28x - 22x = 350800 - (13900 x 22) 6x = 45000 x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Q7. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:
Explaination / Solution:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).  20 - x x 100 = 25 x 2000 - 100x = 25x

125x = 2000 x = 16.

Q8. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
Explaination / Solution:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295. Required percentage = 295 x 100 % = 1475 % = 70% (approximately). 420 21

Q9. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
Explaination / Solution:

Let their marks be (x + 9) and x.

 Then, x + 9 = 56 (x + 9 + x) 100 25(x + 9) = 14(2x + 9) 3x = 99 x = 33

So, their marks are 42 and 33.

Q10. A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
Explaination / Solution:

Suppose originally he had x apples.

Then, (100 - 40)% of x = 420. 60 x x = 420 100 x = 420 x 100 = 700. 60