# Download Puzzle Data Interpretation Based on Time and Distance PDF for SBI PO, IBPS PO, IBPS RRB PO, LIC AAO, LIC ADO 2020

Diretions : Study the following information carefully and answer the questions given beside.

Ram goes to a hill station by car. While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol but while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol. He goes from the point A to the point B. The total distance between point A and point B is 525 km in which the total distance travelled by him uphill is 2.5 times of the total distance travelled by him downhill and the total distance travelled by him on the plane surface is 140 km. While coming back from the point B to point A, he saves 15 litres of petrol and the consumption of petrol is normal on plane surface.
Important for :
1
What is the mileage of the car on downhill?
» Explain it
E
 2x litre per kilometre = 2 litre per kilometre 33
= 1 litre per 16.5 kilometres

Hence, option E is correct.

Common explanation :

Let the normal consumption of petrol = 4x litres per kilometre

While going Uphill, consumption of petrol = 5x litres per km (While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol)

While going downhill, consumption of petrol = 2x litres per kilometre (while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol)

The total distance between A and B = 525 KM

Let the total distance travelled by him downhill = d km then, the total distance travelled by him uphill = 2.5d km

According to the question,

2.5d + d + 140 = 525

 By solving, d = 385 = 110 km 3.5

Total uphill distance = 110 × 2.5 = 275 km

Total downhill distance = 110 km

While going from the Point A to point B, the car will consume total petrol of

5x × 275 + 2x × 110 + 4x × 140 litres = 2155x litres ...........(i)

While coming from point B to point A, plane surface will be plane only but downhill distance will become uphill and the uphill distance will become downhill then plane surface distance = 110 km

Downhill distance = 275 km, uphill distance = 110 km

The total consumption of petrol while coming back from the point B to point A = 2X × 275 + 5X × 110 + 4X × 140 = 1660x litres .................... (II)

According to the question, while coming back from the point B to point A, he saves 7 litres of petrol

It means, 2155x – 1660x = 15 litres

 x = 15 = 1 495 33

2
If point A to point B were a plane surface then how many litres of petrol he would have consumed more while going and coming back?
» Explain it
D
The total petrol consumption while going and coming back

 = 2155 + 1660 = 3815 litres 33 33 33

The mileage of car on the plane surface = 4x litre per km
 = 4 × 1 litre per kilometre 33

While going and coming back, the total distance = 525 × 2 = 1050 km
 1 km = 4 litre 33

 1050 km = 1050 × 4 litre = 4200 litres 33 33

 Reqd. difference = 4200 – 3815 = 385 litres = 11.67 litres 33 33 33

Hence, option D is correct.

Common explanation :

Let the normal consumption of petrol = 4x litres per kilometre

While going Uphill, consumption of petrol = 5x litres per km (While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol)

While going downhill, consumption of petrol = 2x litres per kilometre (while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol)

The total distance between A and B = 525 KM

Let the total distance travelled by him downhill = d km then, the total distance travelled by him uphill = 2.5d km

According to the question,

2.5d + d + 140 = 525

 By solving, d = 385 = 110 km 3.5

Total uphill distance = 110 × 2.5 = 275 km

Total downhill distance = 110 km

While going from the Point A to point B, the car will consume total petrol of

5x × 275 + 2x × 110 + 4x × 140 litres = 2155x litres ...........(i)

While coming from point B to point A, plane surface will be plane only but downhill distance will become uphill and the uphill distance will become downhill then plane surface distance = 110 km

Downhill distance = 275 km, uphill distance = 110 km

The total consumption of petrol while coming back from the point B to point A = 2X × 275 + 5X × 110 + 4X × 140 = 1660x litres .................... (II)

According to the question, while coming back from the point B to point A, he saves 7 litres of petrol

It means, 2155x – 1660x = 15 litres

 x = 15 = 1 495 33

3
The quantity (in litres) of petrol consumed for the entire journey (from point A to point B and from point B to point A) is
» Explain it
E
The total petrol consumption while going and coming back

 = 2155 + 1660 = 3815 litres = 115.6 litres 33 33 33

Hence, option E is correct.

Common explanation :

Let the normal consumption of petrol = 4x litres per kilometre

While going Uphill, consumption of petrol = 5x litres per km (While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol)

While going downhill, consumption of petrol = 2x litres per kilometre (while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol)

The total distance between A and B = 525 KM

Let the total distance travelled by him downhill = d km then, the total distance travelled by him uphill = 2.5d km

According to the question,

2.5d + d + 140 = 525

 By solving, d = 385 = 110 km 3.5

Total uphill distance = 110 × 2.5 = 275 km

Total downhill distance = 110 km

While going from the Point A to point B, the car will consume total petrol of

5x × 275 + 2x × 110 + 4x × 140 litres = 2155x litres ...........(i)

While coming from point B to point A, plane surface will be plane only but downhill distance will become uphill and the uphill distance will become downhill then plane surface distance = 110 km

Downhill distance = 275 km, uphill distance = 110 km

The total consumption of petrol while coming back from the point B to point A = 2X × 275 + 5X × 110 + 4X × 140 = 1660x litres .................... (II)

According to the question, while coming back from the point B to point A, he saves 7 litres of petrol

It means, 2155x – 1660x = 15 litres

 x = 15 = 1 495 33

4
If the speed of car is 55 km per hour on the plane surface and while going uphill, the car’s speed was decreased by 25% of the normal speed and while going downhill the car’s speed was increased by 50% of the normal speed then approximately how much time he would have taken during the entire journey? (if he returns immediately from point B to point A)
» Explain it
B
While going from Point A to point B, Distance =  275 km uphill + 110 km downhill + 140 km on the place surface -------- (i)

While coming back from the point B to point A

Distance = 140 km on the plane surface + 110 km uphill + 275 km downhill ------- (ii)

The total distance while going and coming back = 280 km on the plane surface + 385 km uphill + 385 km downhill (by adding equation (i) and equation (ii))

On the plane surface, the speed of car = 55 km per hr

On uphill, the speed of the car = 75% of 55 = 41.25 km per hour

On downhill, the speed of the car = 150% of 55 = 82.50 km per hour

 The total time taken = 280 + 385 + 385 55 41.25 82.50

= 5.09 + 9.33 + 4.67 = 19.09 hours approximately

Hence, option B is correct.

Common explanation :

Let the normal consumption of petrol = 4x litres per kilometre

While going Uphill, consumption of petrol = 5x litres per km (While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol)

While going downhill, consumption of petrol = 2x litres per kilometre (while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol)

The total distance between A and B = 525 KM

Let the total distance travelled by him downhill = d km then, the total distance travelled by him uphill = 2.5d km

According to the question,

2.5d + d + 140 = 525

 By solving, d = 385 = 110 km 3.5

Total uphill distance = 110 × 2.5 = 275 km

Total downhill distance = 110 km

While going from the Point A to point B, the car will consume total petrol of

5x × 275 + 2x × 110 + 4x × 140 litres = 2155x litres ...........(i)

While coming from point B to point A, plane surface will be plane only but downhill distance will become uphill and the uphill distance will become downhill then plane surface distance = 110 km

Downhill distance = 275 km, uphill distance = 110 km

The total consumption of petrol while coming back from the point B to point A = 2X × 275 + 5X × 110 + 4X × 140 = 1660x litres .................... (II)

According to the question, while coming back from the point B to point A, he saves 7 litres of petrol

It means, 2155x – 1660x = 15 litres

 x = 15 = 1 495 33

5
What is the difference between the mileage of car on downhill and that on uphill?
» Explain it
C
The required difference = 5x – 2x = 3x = 3/33 = 1/11 litres per kilometres = 1 litres per 11 kilometres

Hence, option C is correct.

Common explanation :

Let the normal consumption of petrol = 4x litres per kilometre

While going Uphill, consumption of petrol = 5x litres per km (While going upwards (uphill) the consumption of petrol was increased by 25% of the normal consumption of petrol)

While going downhill, consumption of petrol = 2x litres per kilometre (while going downwards (downhill) the consumption of petrol was decreased by 50% of the normal consumption of petrol)

The total distance between A and B = 525 KM

Let the total distance travelled by him downhill = d km then, the total distance travelled by him uphill = 2.5d km

According to the question,

2.5d + d + 140 = 525

 By solving, d = 385 = 110 km 3.5

Total uphill distance = 110 × 2.5 = 275 km

Total downhill distance = 110 km

While going from the Point A to point B, the car will consume total petrol of

5x × 275 + 2x × 110 + 4x × 140 litres = 2155x litres ...........(i)

While coming from point B to point A, plane surface will be plane only but downhill distance will become uphill and the uphill distance will become downhill then plane surface distance = 110 km

Downhill distance = 275 km, uphill distance = 110 km

The total consumption of petrol while coming back from the point B to point A = 2X × 275 + 5X × 110 + 4X × 140 = 1660x litres .................... (II)

According to the question, while coming back from the point B to point A, he saves 7 litres of petrol

It means, 2155x – 1660x = 15 litres

 x = 15 = 1 495 33

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Data Interpretation is an important part in all bank exams. Here we at Smartkeeda.com we will learn a New Pattern Based DI, Which is Called Caselet DI, These days Caselet DI freequenty asked in all major Bank exams SBI PO Pre, SBI Clerk. It has alot chance to appear in SBI PO 2019 and SBI Clerk 2019.

We provide you data interpretation quiz with answers and explanation.

Aspirants preparing for SBI PO exams for the year 2019 can practice these questions. Let us now understand, what exactly is a Caselet DI. Data interpretation normally consists of questions involving pie charts, bar graphs, line graphs, radar graphs pr table with the required information for solving the questions. In a paragraph type data interpretation question, a set of information is provided in a paragraph form. It doesn’t consist of any charts or tables. You have to read the given information carefully and draw a suitable table/chart listing out all the given data to answer the questions.

So, First of all, whenever exams authorities comes up with new kind of questions, students raises one query, "How to Solve New Pattern Based Caselet Data Interpretation". So first we discuss
How to solve caselets/paragraph/Info Chart data interpretation questions?

Given Below steps will help you all while solving questions of Caselet or paragraph based Data Interpretation questions in SBI PO, SBI Clerk Exams. Data Interpretation for SBI PO Pre, SBI Clerk 2019 with PDF.

I. While Solving caselet involves a thorough understanding of the subject matter of the passage given. Read the paragraph with utmost care and analyze what the question demands. Basically be clear on what is given and what is asked.

II. Try to underline all the important points in a caselet while reading it. You can always use symbols in place of names of persons, places etc to make your work a lot easier. Focus only on useful data and don’t do unnecessary approximations just to simplify the task.

III. With all the important information in your hand, try to represent the data in graphical or tabular form. Represent the data in a pie chart, bar graph, table etc depending upon the feasibility and the motive of the question.

IV. Read the figures closely. You can use options and approximations to avoid tedious and lengthy calculations. If the question asks for relative values, there is no need for find the accurate values. Use approximation but never over-approximate.

Sks Here are the some points you need to remember:

I. Improve your calculation speed: Caselets being calculation intensive, you have to be good with your calculations. Learn speed math techniques and practice them regularly. The more you practice the faster you become in calculations. After a considerable amount of practice, you can do the calculations in your mind. This is going to save a lot of your precious time in exam conditions. So build a intuitive number sense.

II. Be thorough with the arithmetic related topics like percentages, interests, ratios and proportions as caselets are often based on these type of concepts. Otherwise you will face a lot off difficulty tackling these questions. Get the basics; understand the difference between growth and growth rate, average growth rate, cumulative average growth rate, market share by volume and market share revenue etc.

III. Practice makes a man perfect. The more you practice these types of questions, the easier the questions will become. Practice questions will improve your speed and accuracy. While attempting these type of questions in exam, always try to choose the questions which will give you the answer with lesser effort, or you will end up spending a significant amount of time on lengthy calculations.
Caselet DI is frequently being asked these days and so we must practice this in the form of quizzes as Smartkeeda Caselt Data Interpretation Quiz for Free of Cost. So, Caselet DI is not new or not even out of way. It is just a mathematical form of English Passage. In Caselet DI, a long paragraph is given and on the basis of that. Candidates can download data interpretation quiz with solutions for banking and other competitive examination.

In the above paragraph of Caselet DI, lots of information is given. You have to read the paragraph carefully and then you have to note down all the key information as short as possible. The given information will let you draw some diagrams such as Venn diagram, tabular chart or any other diagram.

The difference between simple DI and Caselet DI is - In simple DI, information is already given in diagrammatical forms but in Caselet DI, you have to draw a diagram on the basis of given information

Before start solving Caselet DI, you must have knowledge of the following 7 key things:

1. How to draw a diagram on the basis of given information.

2. Which diagram is the need of the question?

3. Knowledge of Venn diagram.

4. Knowledge of the relationship between fractions and their percentage forms.

5. Knowledge of simplification and approximation.

6. If the question belongs to CI & SI or Profit & Loss or Speed, time and distance then you must have knowledge of basic formulae of these topics.

7. Must have knowledge of Ratio and Proportion.

So, in the above information chart or Caselet Chart or Passage Chart you will see all the rules are following properly. While attempting Caselet DI you can even save your time in exam only one condition if you do practice a lot which trough you can create Image or differentiate the data and After that you can answer easily.'

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