CEED 2020 Part A answers with detailed solutions

CEED 2020 question paper for reference

Part B solutions will follow this Part A answers in another post.

Part B sketches CEED 2020 paper

Detailed explanations to some of the design aptitude questions of this paper are available in the below set of videos

Answers

1)  14

The object shown in the front view is symmetry in the other four sides (front, right, back and left). However, seeing the top view it is clear that other than shape 2, all other shapes are either cylindrical or spherical.

Shapes 1 and 4 are spherical kind. 

Shapes 3, 5 are cylindrical while shape 6 is a cut hole inside 5.

No. of surfaces due to shape 1 = 1

No. of surfaces due to shape 4 = 1

No. of surfaces due to shape 3 = 1 (not considering the top view surfaces)

No. of surfaces due to shape 5 = 1 (not considering the bottom view surface)

No. of surfaces due to shape 6 = 1 

No. of surfaces due to shape 2 = 4 (considering only four sides - front, right, back and left)

semi total = 9

As shown in the second figure (you can see the sides mentioned), in top view, two surfaces shall be considered denoted as 1 and 2 (this numbering 1 and 2 is different in the first image).

Now, finally in the third image, you can count the no. of surfaces from the bottom as 3. 

Total = 9+2+3 = 14

To get familiar with this type of questions, refer to the posts on
'counting no. of surface',

'3d manipulation', and

'tessellation' posts in this blog Part A resource pages.

2) 6

As can be seen, the no. of ink marks is total 9. Ignoring the initial ink drop, the total no. of impressions will be 8. 

Now, we need to find the no. of revolutions (rev) of the front wheel such that the combined no. of ink marks due to the front and the back wheel counts to 8. 

Let's say the diameter of the front wheel = x

diameter of the back wheel = 1.5x (as given in ques)

this means, for one complete rev of the back wheel, the front wheel would have done 1.5 rev

Let's consider the front wheel rotates 5 times, then the no. of rev of the back wheel will be 5/1.5 = 3.33, 

Combined revs (5+3.33) is more than 8. However, note that when the small wheel goes over the ink at the start, the back wheel is at a distance from the front and so it takes some revs to reach that ink spot. Considering a safe distance, we need to add one more rev to front.

Trying 6 revs for front wheel, will cause the back wheel to revolve 6/1.5 = 4, totaling to 10, which is the most feasible solution. 

3) 20

As shown in the first image, the surfaces numbered 1,2,3 and 4 add 4 surfaces. This is extended to make a U shape and hence the whole four surfaces will get enlarged, joining two X shapes. Also, observe that X has four sides, denoted as a,b,c,d. This when extruded and for two of such parts (as shown in X1 image) will have 4+4 = 8 surfaces. (Total, tripled). Now, the two bottom surfaces of these two extrusions will add another 2 surfaces. 

Total surface for X1 is thus 4 + 8 + 2 = 14

Similarly, for Y, we just need to find the no. of sides in Y. Two of Y shapes will be joined by an U shaped extrusion of Y. So, the trick is find the no. of sides of Y, thripple it and add 2 more surfaces. 

No. of sides of Y is 6 (as shown below)

Total surfaces = 6*3 + 2 = 20 

4) 32

No. of visible triangles, including the small triangle = 26 (shown as yellow in the below image)

No. of bigger triangles as seen in the next two images = 3+3 = 6

Total no. of triangles = 26+6 = 32

5) 18

Shown in the below figure, one possible route taking 18 grid steps 

6) 8

A good way to start is to watch from left to right or the reverse and keep comparing both the images to identify the differences as shown in the image below.

Similar and complete Part A resources 

7) 1690

As shown in the below image (only cross-section in 2D was shown), the best possible way was to stack two such objects denoted by 1 and 2. Then 3,4 will be stacked as shown by utilizing the space, note that shapes 5, 6 will also be stacked on the remaining two sides (assume a rectangular box with 1,2,3,4 are the front, right, back and left sides, then 5 and 6 shall be the top and bottom of the box). The sides will have size 10 and 13 as shown in the figure.

Volume = 10*13*13 = 1690

8) 24

Below, I've given a detailed answer to this.

In the first image below, I've marked the surfaces with numbers for your easy understanding. 

As shown, surface 1 spans around the solid as it is continuous. You can compare the surface numbering in both the two views of the solid shown in the first image below. (count = 1)

Surface 2 is a semi-spherical hollow and it counts for 1 surface. (count = 1 +1)

Number 3 is a continuous surface that has two D slots as shown in both the views. Note that there is are no sharp edges except the two D slots and hence we consider that as 1 surface. (count = 2+1)

Like 3, you can visualize three more surfaces around the object (two surfaces numbered 4 and 5 are shown while another surface can be assumed to be at the backside of both the views) (count = 3+3)

In the second image, number 6 is again a conical continuous surface and should be counted as 1 (count = 6+1)

Like no. 7, numbers 8,9 and 10 are all continuous all around the circumference and hence should be taken as 1 surface each. (count = 7 +4)

cut for number 11 is bit tricky as you can see that there are breaks and it is not continuous. I could show three such breaks (11a, 11b, 11c) while the fourth is behind the view. So, total 4 surfaces (count = 11 +4)

Surface 12 is again continuous (count = 15 +1)

Now, the only surfaces leftover are the D type slots on each (four) projected surfaces. 

As shown in below image (2D view), the D shape consists of one straight and a circular surface (shown in two different colors) and so each D slot will contribute to 2 surfaces. For four such slots, the total no. will be 8. (Count = 16+8 = 24)

9) A, B, D

What I would do is check the position of the circle and square shapes with respect to a corner as shown in the below image. With this, I can say that C is missing the pattern. 

10) A, D

I started with the first pattern of the circle as highlighted in the below image and cross-checking that with the options, I can see that B is not matching as shown in the second image. 

Likewise, I randomly considered the highlighted (in yellow) pattern in the ques and checked for mismatch. C was thus eliminated. 

Trying with some other random pattern will give further confirmation of the right options.

Find non verbal and reasoning related materials here

11) C

By Y axis, one possible way to check is by the movement of the second part (away from us) as shown below. But the pin restricts the movement of the board due to slanted. 

Another important thing to observe is that - if the two edges (highlighted as 1 and 2 in the second image) are parallel or are converging in the Y-direction, then it is not easy to move the parts due to the pin. 

In option B, the two surfaces are parallel and on top of that, the rhombus pin at the interface prevents the movement of the part. 

In option C, as shown below, the two surfaces are diverging and also the pi is just at the corner of the two surface interfaces. 

In option D, Though the two surfaces are diverging, because of the presence of a little projection at the interface as highlighted in yellow in the image, the movement of the parts is restricted in Y direction.

12) B

By observation,

Further, I've highlighted the locations where other patterns are losing the circularity.

13) C, D

I will give a detailed answer below.

To answer this question, you need to understand the physics behind the operation of the spray as given in the question. Let me explain how this functions. In the first image below, when the air was blown at the shown position, the blow of air at that intersecting junction causes the air to flow at high velocity. In physics, a high velocity happens at the cost of pressure drop, which means low pressure. At the same time, the atmospheric pressure at the surface shown with green arrows will always try to push the liquid. This simultaneous action of air pushing and pressure reduction causes the liquid to be sucked from the narrow tube, hence the spraying action.

Now that you understood the physics, let's check the options. In option A, when low pressure gets created at 1, the liquid might try to be sucked up till 1. However, since the narrow tube is connected to another tube and is open to the atmosphere at 2, the low pressure at 1 sucks air instead of liquid (of higher density) from 2 that is open to the atmosphere. Hence the spraying action will not happen. 

In option B, even though 2 is connected inside the liquid and is open to the atmosphere at one side (atmospheric pressure acts at 2), the low pressure at 1 will cause the liquid to be sucked up. However, the inside portion of the bottle is enclosed and is not open to the atmosphere. So, the air entrapped inside it will try to create a vacuum as the liquid level tries to go down (due to an increase in space for air inside the bottle). This negative vacuum will be strong enough to hold the liquid that the small pressure drop at 1 will not have any action or maybe a minor action. 

In option C, the inner top surface of the liquid in the bottle is open to the atmosphere by way od opening at 2. This always keeps the air pressure there at atmospheric. When negative pressure happens at 1, the liquid will be easily sucked up due to higher pressure on the liquid inside the bottle. 

In option D, since the two narrow tubes are independent unlike option B, the atmospheric pressure at 2 tries to push the liquid inside any low pressure created inside the bottle will be compensated by the flow of liquid from the tube in 2. Remember, the atmospheric action on 2 will continue until the liquid level drops to the bottom level of the smaller test tube bottom. After that, the bottle gets open to the atmosphere due to opening at 2 and hence the effect of atmospheric pressure will be higher than before and hence the liquid keeps sucked until the level of liquid goes below the level of the vertical tube.  

14) B

Forming is the usual process employed for sheet metal. 

Casting and blow molding are mostly for solids but not for sheet metals. Brazing is a post-casting process.

Materials and manufacturing process

15) B, C

The step as highlighted at position 1 in the below image is missing in the pattern shown in option A. In option D, the option shown as 2 is not correct. B and C can be obtained by rearranging the stamping. 

16) All

Fold the post in half along the longer side of 12cm to have a measure of 6cm too. 

Now, using the post card as a ruler/scale on the page of the notebook, draw a line with 7cm side of the postcard. Continue drawing the line by keeping the folded half of 6cm to make a straight line of 6+7cm = 13cm. Using the compass, draw a circle keeping the two edges at the end of this line to get a circle of 13cm radius.

Draw a straight line using postcard as a ruler along 12cm side. Now, turn the postcard 90 degrees and align the 7cm side of the post card along the already drawn line of 12cm. Mark the 7cm. Now erase this 7cm to get a 5cm line. Using the compass and keeping a radius of 5cm, cut arcs to get the other vertex and joining lines from each end of the first line to this new point to get an equilateral triangle. 

Use the half size of the postcard (6cm) to draw a square of side 6cm. 

Likewise, use the 5cm measure from B and if needed the compass to make a hexagon of side 10cm

17) C, D

Use the criteria shown in the below image.

For the pattern in row 1, two solid inverted U cups followed by one non-solid inverted U cup.   

The distance 2 and distance 3 of the two extra solid cups. Consider this as a first check to eliminate the nonmatching options. Cross-check with the options to see which option matches these three criteria. 

Option A is not meeting the criteria 1 as the pattern of two solid cups followed by one open cup is missing. Note that option A is to be cross-checked by rotating R 180 degrees. 

In option B, criteria 1 is matching, but criteria 2 and 3 are not matching. 

Option C is a rotated view (180 degrees) of R and is matching all criteria.

Option D is an as-it-as view of R with all the criteria matching. 

As we have eliminated two options, we can now check all other patterns within the chosen options to see if there are any mismatches. Fortunately, C and D fits.

18) D

The quote says that there can be two natures of worlds - static and changing. Probability is the relation applicable for static worlds while causality signifies the changes in the probability for the changing world. 

So, option A is not correct.

Option B is wrong as changes in the world is predicted by causality, which tells the changes in the probability

C is never related to the quote.

D is closely related since Probability talks about a limited application (for static world) while causality is for a bigger picture (for all kinds of probable worlds)

19) C

As shown in the below picture, the rays should expand (diverge) inside the house, indicated by yellow arrows. Also, the small gaps of the mesh/curtain should allow the light of the appropriate width as highlighted by the red arrow. 

In option A, light is all along the front body of Suman which is wrong as shown in the below picture. 

In option B, light is not at all present where it meant to be as shown in the below picture. 

Again, option D is wrong, as light is spanning above and below the arrows as shown for the third option above. 

20) C

For easy understanding, I've made the below image. As can be seen, the scenario can be interpreted as if the door is hinged to the left and the opening end of the door is at the right. The door opens towards us in the direction indicated by the red arrow.

With this scenario, we tend to use our right hand to handle the key while the left hand shall be used to operate the lever. Now, with the right hand, it is custom to open the door by turning the key anti-clockwise (as in most of the lock cases). Note that to lock the key, the key has to be rotated clockwise as well. To operate the lever, it is ergonomically easy to turn it down. Turning it up is kind of painful. Option C seems to be ok. 

21) C

The direction of the light rays from the two light sources and the reflector function are shown in the below image. Light from P onto the two products acts as shown by blue arrows. The reflector is aligned almost perpendicular to P and so the light gets deflected along with green arrows (showing approx. to give an idea.). Unfortunately, like R, S is also going to be useless as the light rays from S is not useful to lit the product as shown in red color. With this configuration, option C shows the right shadow direction and texture on the products.

22) C

I kept observing the distance of the cut from the left and verified the approx. distance with the options as shown in the below picture. Likewise, you can visualize the distance of the cuts from the right edge too.

23) C

I've denoted the cuts with numbers as shown below. Among the option, C matched the numbered shapes as shown in the second image.

Options A and D have more than 7 shapes. In option B, pattern 2 is repeating twice, which is wrong.

24) C

The wheel seems to rotate in the counterclockwise direction with blue in the center.

25) A

I've explained the logic in the below picture. The numbering shows the order by which the pattens need to be checked. I've only considered the yellow square initially to see where it lands at the question mark of 9th part.

In each step (number), the yellow square is moving one step in anti-clockwise direction and after three steps, it is changing to green color in the fourth step as shown. Following this pattern, the green square in the 8th pattern will change color to yellow and it will occupy the position as highlighted by the arrow in the 8th numbered pattern. 

26) A

To make the observation easy, what I did is assign number 1 to the sticker facing up. The stickers facing down can be taken as 0. Now, as shown in the image, it is now easy to spot the patterns with up (no. 1s) and the rest are patterns downwards.

27) C

A dice is numbered such that the sum of the numbers on the opposite faces add up to 7 always. 

In the below image, the construction is shown. The pairs of numbers are 

(1,6)

(2,5)

(3,4)

Now, when the dice as shown in the question rolls to B, the base face will have the number 6. So, it's the opposite face (the top face) will be 1 as per the combination listed above. number 3 will be the front face. Opposite to face with 2, 5 will be present. When the dice rolls to C, the face with number 5 will be the bottom while 2 will be the top now. At the same time, 6 will be to the left face and hence 1 will be on the right face at this position. When the dice rolls to D, 1 will become the bottom face and 6 becomes the top face. while number 3 will continue to be the front face. when the dice rolls again to E, face with number 3 will bottom and hence face with the number (7-3 =4) will be the top. 

28) B

All you need to do is to visualize to make path (shortest path) for all the options and see which option's path matches the red line in the question puzzle with incomplete black lines. I've drawn all the options for your understanding. Hope this is clear.

29) D

This is how I interpreted the symbolic representations.

1. wrongs cannot be right

2. pen is mightier than a sword

3. Time is money

4. Teamwork is better than individual work

5. A happy life/person needs/means lesser medicines/sick

30) No answer

The question is incomplete in the paper

31) B

For a detailed list and collection of paintings and painters and more about Indian culture, refer to the CEED Part A resource page in this blog 

32) D

By careful observation and comparison.

Option:

A: bottom teeth are missing

B: Front big tooth is missing

C: Tail is not aligned properly

D: All are intact

For such question on picture to picture comparison, refer to the CEED Part A resource page in this blog

33) B

Here's - list of Indian logos 

34)  

I'm not getting this :D

35) C

By careful observation, it is clear that C is the right option. In option D, the fact that the lines on the two sheets (as highlighted by arrow)are straight - disqualifies this option.

36) D

By observing the notion of symmetry, ends, and curves, D seems a fit. 

37) C

I've joined the dark dots to see alphabets as shown in the below image. However, they are upside down (inverted). 

In the second image, I've rotated the whole image 180 degrees (flipped along the horizontal to get the image for a clear view of the alphabets. As you can see, the alphabets are: 

O, R, A, N, G, E

38) D

The answer will be simple to figure out if you can understand the logic of where to start

I've just considered the starting first curve as highlighted in the first image. For this curve to happen, the radius is shown and along the slant angle of the starting was approximately shown. This is purely based on intuition. 

Now, observe the options and given the direction of rotation of the objects, spot the corner/vertex which will be the point at which the object turns. I've marked that for all the options in red. In option A, the turn radius is too small. Options B and C are also small. Moreover, the height of the start point in B is higher from the ground than that shown in the given question. C radius is small enough to make the curve shown in the question. D suits the purpose.

39) A

The pattern goes like the numbering shown in the image.

When color a and color b mix results in color a1

when color b and color 6 mixes give color a6. 

Likewise, the question mark should be c3, c4 and d3, d4 as shown.

Since c,3 and d,4 are shades of magenta, all the four squares will be shades of magenta. Option A is the only choice.

40) D

This is straight forward and simple question. Anyway, the below rotation indication shows the turning of the object.

41) B

By opposite side, they mean the backside of the robot. This is more like a mirror image.

In the image below, the first criterion that I chose in eliminating the wrong options is the ears of the robot as highlighted in the first row. Given this position, in the back view, only a partial view of the ears will be visible. So, option A and D gets eliminated.

The second criterion to screen between options B and C is the position of the robot legs as highlighted in the second row. Clearly, option D got eliminated.

Step to step animation process - detailed explanation


Further: Complete list of solutions to previous years CEED and UCEED papers