Search Previous Year Questions

Given:

A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is \(35 \dfrac{7}{17}\). 

Formula Used:

average = sum of observations/number of observation

average of n consecutive number = [n(n + 1)/2]n = (n + 1)/2

Calculation:

Since the average of the remaining numbers is \(35 \dfrac{7}{17}\) 

To get an average greater than 35, the value of n must be near 69.

Let initially number of terms be 69 and the number which was erased be 'x'

⇒ (69 × 35 - x)/68 = 602/17

⇒ 2415 - x = 2408

⇒ x = 7

∴ The number erased was 7

 

Now,

A pair must walk together at the speed of the person who walks slowly.

As, A takes the shortest time, so each friend will walk with A.

Therefore:

A crossing the bridge with B and then retuning = 1 + 2 minutes = 3 minutes

A crossing the bridge with C and then retuning = 1 + 8 minutes = 9 minutes

A crossing the bridge with D = 11 minutes

A won’t return as none of the friend is left.

So, the minimum time required to cross the bridge by all the four people = (3 + 9 + 11) minute

= 23 minutes

Hence, ’23 minutes’ is the correct answer.

Given:

Percentage of illiterate adults = 40.1% ≈ 40% 

Percentage of literate children = 85.1% ≈ 85% 

The ratio of the adults to that of children = 2 ∶ 3 

Calculations:

Let the total adult be 2x and children be 3x 

Total population = 2x + 3x = 5x

Total illiterate adults = 2x × 40/100 = 4x/5 

Total literate adults = 2x – 4x/5 = 6x/5

Total literate children = 3x × 85/100 = 51x/20

Total literate = (6x/5) + (51x/20)

⇒ (24x + 51x)/20 = 75x/20 

Percent of the literate population = ((75x/20) ÷ 5x) × 100 = 75% 

∴ Percent of the literate population is 75%

  

Let the total adult be 200 and children be 300

Total population = 200 + 300 = 500

Total illiterate adults = 200 × 40/100 = 80 

Total literate adults = 200 – 80 = 120

Total literate children = 300 × 85/100 = 255

Total literate = 120 + 255 = 375

Percent of the literate population = (375/500) × 100 = 75% 

∴ Percent of the literate population is 75%

Given:

A runs \(1 \dfrac{2}{3}\) times as fast as B 

A gives B a start of 80 m 

Concept used:

Time = Distance/Speed 

Calculations:

Let the speed of B be x 

So, speed of A = 5x/3 

Let the total distance be 'a' m 

So, time taken by A to complete the total distance = a/(5x/3) = 3a/5x 

A gives B a start of 80 m 

Required distance for B = a – 80 

Time taken by B to complete the distance = (a – 80)/x 

 A and B might reach at same time 

∴ Time taken by A to complete the total distance = Time taken by B to complete the distance

⇒ 3a/5x = (a – 80)/x 

⇒ 3a = 5x – 400 

⇒ 2a = 400 

⇒ a = 200 

∴ The winning post must be 200 m far so that A and B might reach it at the same time

  

Let the speed of A and B be 5x m/s and 3x m/s respectively 

A gives B a start of 80 m i.e. distance between them = 80 m

relative speed = (5x - 3x) = 2x m/s

time required = 80/2x = 40/x

distance traveled by A = 5x × 40/x = 200 m 

∴ The winning post must be 200 m far so that A and B might reach it at the same time

Let the age of the person’s mother be ‘a’ units.

So, age of the person = (2/5)a

After 8 years, age of the person’s mother = ‘a + 8’ units.

So age of the person after 8 years = (a + 8)/2

According to question:

(2/5)a + 8 = (a + 8)/2

=> (2a + 40)/5 = (a + 8)/2

=> 4a + 80 = 5a + 40

=> a = 40

So, the present age of his mother is 40 years.

Hence, ’40 years’ is the correct answer.

Given:

The average speed of Mr. Kumar = 48 km/h

The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance 

Concept used:

Time = Distance/Speed 

Calculations:

Let the total distance be x 

60% of the distance = x × 60/100 = 3x/5 

Time taken by Mr. Kumar to cover 3x/5 distance = \(\frac{{3x}}{{5 \times 48}}\) = x/80

Remaining distance = x – 3x/5 = 2x/5 

Time taken by Mr. Kumar to cover 2x/5 distance = \(\frac{{2x}}{{5 \times 48}}\) = x/120 

According to the question,

⇒ (x/80) – (x/120) = 10/60 

⇒ (3x – 2x)/240 = 10/60 

⇒ x = 40 

∴ The distance of office is 40 kms

Given:

A can fill the cistern = 37.5 minutes = 75/2 minutes

B can fill the cistern = 45 minutes

Concept used:

Total work = Efficiency of a person × Time

Calculation:

Taking the L.C.M. of 75/2 & 45 = 225 (which is total work)

Work done by A in 75/2 minutes = 225

Efficiency of A in 1 minute = (225 × 2)/75 = 6 

Work done by B in 45 minutes = 225

Efficiency of B in 1 minute = 225/45 = 5

Let the turned off time of B be x minutes.

Efficiency of A in 1 minute × Time + Efficiency of B in 1 minute × Time = Total work 

⇒ 6 × 30 + 5 × x = 225

⇒ 180 + 5x = 225

⇒ 5x = 225 – 180 

⇒ x = 45/5

⇒ x = 9 minutes

∴ B is turned off after 9 minutes.

 

Since pipe A has opened for 30 minutes, part of cistern filled = 30/37.5 = 4/5

part of cistern filled by B = (1 - 4/5) = 1/5

time taken by pipe B = 1/5 × 45 = 9 minutes

∴ B is turned off after 9 minutes.

The logic is:

 

Similarly,

Hence, ‘354#’ is the correct answer$.

Given:

Height of rain water = 5 cm

Area of ground = 1.5 hectares = 1.5 × 10⁸ cm²

Concept used:

Volume = Area of base × Height

1 hectare = 100000000 cm² = 10⁸ cm²

1 cm = 1/100 m

Calculation:

Volume of water that falls on the ground = 1.5 × 10⁸ × 5

⇒ 75 × 10⁷ cm³

⇒ (75 × 10⁷)/10⁶ m³

⇒ 75 × 10 m³

⇒ 750 cubic meter

∴ The volume of water that falls on 1.5 hectares of ground is 750 cubic meter.

Persons: A, B, C, D, E, F, G and H

1. A sits neither opposite to D nor to H but is sitting in between C and G. 

2. B sits neither opposite to A nor to G but is sitting in between F and D 

3. B and H belong to team X and D sits opposite to E 

As B and H belong to team X, so is C and G.

Remaining F, E, A and D belong to team Y.

Hence, ‘B, H, C and G’ is the correct answer.