EPFO Assistant Prelims Mock Test - 8 - Bank Exams MCQ

From the following explanation we get to know that 38 is fifth to the left of 49 in step 4.
Step 4: 72 60 38 16 88 94 29 49 65 76
Hence option B is the correct answer

Complete Machine Process:
Input: 88 59 28 94 37 75 15 64 71 48
Step 1: 16 88 59 94 37 75 64 71 48 29
Step 2: 38 16 88 59 94 75 64 71 29 49
Step 3: 60 38 16 88 94 75 71 29 49 65
Step 4: 72 60 38 16 88 94 29 49 65 76
Step 5: 89 72 60 38 16 29 49 65 76 95

Common Explanation:
Change happening from Input to Step 1:

Reference:
Input: 62 97 38 74 55 12 86 45 68 22
Step 1: 13 62 97 38 74 55 86 45 68 23

Inference:
In every step, two numbers are changing their positions. In input, the lowest number which is 15, is first added by 1 and getting placed at extreme left end. Similarly, the second lowest number among all, which is 22 is also getting increased by 1 and is getting placed at the extreme right end of the input to give us Step 1.

Given Input: 88 59 28 94 37 75 15 64 71 48
Step1: 16 88 59 94 37 75 64 71 48 29
 
Reference
Step 1: 13 62 97 38 74 55 86 45 68 23
Step 2: 39 13 62 97 74 55 86 68 23 46

Inference:
In this step too, the 3rd lowest number among all is first getting increased by 1 and is placed at the extreme left end of Step 1 and so is the case with the 4th lowest number. 

Step1: 16 88 59 94 37 75 64 71 48 29
Step2: 38 16 88 59 94 75 64 71 29 49
 
And the same process continues till we get all the numbers arranged in similar fashion.

The complete Machine process is as follows:
Input:  88 59 28 94 37 75 15 64 71 48
Step1: 16 88 59 94 37 75 64 71 48 29
Step2: 38 16 88 59 94 75 64 71 29 49
Step3: 60 38 16 88 94 75 71 29 49 65
Step4: 72 60 38 16 88 94 29 49 65 76
Step5: 89 72 60 38 16 29 49 65 76 95

Clearly, there are two numbers between '16' which is 4th to the left of 76 and 72 in step 5.
Step 5: 89 72 60 38 16 29 49 65 76 95
Hence, option B is correct.

Complete Machine Process:
Input: 88 59 28 94 37 75 15 64 71 48
Step 1: 16 88 59 94 37 75 64 71 48 29
Step 2: 38 16 88 59 94 75 64 71 29 49
Step 3: 60 38 16 88 94 75 71 29 49 65
Step 4: 72 60 38 16 88 94 29 49 65 76
Step 5: 89 72 60 38 16 29 49 65 76 95

Common Explanation:
Change happening from Input to Step 1:

Reference:
Input: 62 97 38 74 55 12 86 45 68 22
Step 1: 13 62 97 38 74 55 86 45 68 23

Inference:
In every step, two numbers are changing their positions. In input, the lowest number which is 15, is first added by 1 and getting placed at extreme left end. Similarly, the second lowest number among all, which is 22 is also getting increased by 1 and is getting placed at the extreme right end of the input to give us Step 1.

Given Input: 88 59 28 94 37 75 15 64 71 48
Step1: 16 88 59 94 37 75 64 71 48 29
 
Reference
Step 1: 13 62 97 38 74 55 86 45 68 23
Step 2: 39 13 62 97 74 55 86 68 23 46

Inference:
In this step too, the 3rd lowest number among all is first getting increased by 1 and is placed at the extreme left end of Step 1 and so is the case with the 4th lowest number. 

Step1: 16 88 59 94 37 75 64 71 48 29
Step2: 38 16 88 59 94 75 64 71 29 49
 
And the same process continues till we get all the numbers arranged in similar fashion.

The complete Machine process is as follows:
Input:  88 59 28 94 37 75 15 64 71 48
Step1: 16 88 59 94 37 75 64 71 48 29
Step2: 38 16 88 59 94 75 64 71 29 49
Step3: 60 38 16 88 94 75 71 29 49 65
Step4: 72 60 38 16 88 94 29 49 65 76
Step5: 89 72 60 38 16 29 49 65 76 95

Rahul’s position 14th from front
John’s Position 17th from last 
Alisha is between them


People between them 48 – (14+17) = 17
Where the midmost is Alisha 
So she has 8 people on either of her side.

So Option A is correct

We have,

• V is four persons senior to Z, who is not a Trainee.
• Only one person is designated between V and W, who is immediately junior to S.

From the above condition, there are three possibilities.

 

Again we have,

• The number of persons senior to W is the same as the number of persons junior to T, who is two persons senior to Y.

From the above condition, Case 1 gets eliminated.

Again we have,

• X is senior to U.
• T is not immediately senior to U.

From the above condition, case 3 gets eliminated. Hence, case 2 shows the final arrangement.

We have,

• V is four persons senior to Z, who is not a Trainee.
• Only one person is designated between V and W, who is immediately junior to S.

From the above condition, there are three possibilities.

 

Again we have,

• The number of persons senior to W is the same as the number of persons junior to T, who is two persons senior to Y.

From the above condition, Case 1 gets eliminated.

Again we have,

• X is senior to U.
• T is not immediately senior to U.

From the above condition, case 3 gets eliminated. Hence, case 2 shows the final arrangement.

Nothing about the relation between political freedom and democracy is mentioned in the statement. So, I does not follow. But II directly follows from the given statement.

2 and 3 are the satements involving the extreme terms of the last 2 premises and form a complementary pair. Thus either 2 or 3 follows

P is the son of Q, Q is the mother of R, R is the sister of S, S is the father of T. P, S and R are the children of Q, and T is the child of S.
P and S are males, whereas R is a female. P is the paternal uncle of T.

He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.

N + A = 10%
N = 8.33%
Hence A = 1.66% Æ 60 days.

The resulting cuboid has dimensions:
ℓ = (6 + 6) cm = 12 cm
b = 6 cm
h = 6 cm
Therefore, surface area of the cuboid
= 2(ℓb + bh + hℓ)
= 2[12 × 6 + 6 × 6 + 6 × 12] cm² = 360 cm² 

Cost ∝ Weight²
Let original weight = 10x
Then, original cost = 100x²k (where k is proportionality constant)
After breakage weight = 1x + 2x + 3x + 4x
After breakage cost = kx²(1 + 4 + 9 + 16) = 30x²k
Now, (100 - 30)x²k = 70,000
Or, 70x²k = 70,000
Or, x²k = 1,000
Hence, original cost = 100x²k
= 100 × 1,000 = Rs. 1,00,000

Average annual exports of Company Z during the given period

From the analysis of graph the exports of Company Z are more than the average annual exports of Company Z (i.e., Rs. 85.71 crores) during the years 1994, 1995, 1996 and 1999, i.e., during 4 of the given years.

Since 66 students get 20 and above marks in Chemistry and out of these 21 students get 30 and above marks, therefore to select top 35 students in Chemistry, the qualifying marks should lie in the range 20-30.

We have 60% of 50 = 
∴ Required number
= No. of students scoring 30 and above marks in Physics
= 32

Excise duty = 41%

First, let's consider the different medal combinations that can be awarded to the 3 winners: (1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE. (2) What if there is a 2-WAY tie? --If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. --If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. --There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total). (3) What if there is a 3-WAY tie? --If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD. --There are no other possible 3-WAY ties. Thus, there are 4 possible medal combinations: (1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G. Now let's determine how many different ways each combination can be distributed. We'll do this by considering four runners: Albert, Bob, Cami, and Dora.
COMBINATION 1: Gold, Silver, Bronze

Therefore, there are different victory circles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist. COMBINATION 2: Gold, Gold, Silver.
Using the same reasoning as for Combination 1, we see that there are 24 different victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. However, it is important to realize that these 24 victory circles must be reduced due to "overcounting." To illustrate this, consider one of the 24 possible Gold-GoldSilver victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a SILVER. Notice that this is the exact same victory circle as the following: Bob is awarded a GOLD. Albert is awarded a GOLD. Cami is awarded a SILVER. Each victory circle has been "overcounted" because we have been counting each different ordering of the two gold medals as a unique victory circle, when, in reality, the two different orderings consist of the exact same victory circle. Thus, the 24 victory circles must be cut in half; there are actually only 12 unique victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. (Note that we did not have to worry about "overcounting" in Combinat ion 1, because each of those 24 possibilit ies was unique.) 
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible victory circles, but only 12 unique victory circles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold.
Here, once again, there are 24 possible victory circles. However, because all three winners are gold medalists, there has been a lot of "overcounting!" How much overcounting? Let's consider one of the 24 possible GoldGold-Gold victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD. 
Notice that this victory circle is exact ly the same as the following victory circles: Albert-GOLD, Cami-GOLD, Bob-GOLD. Bob-GOLD, Albert-GOLD, Cami-GOLD. Bob-GOLD, Cami-GOLD, Albert-GOLD. CamiGOLD, Albert-GOLD, Bob-GOLD. Cami-GOLD, Bob-GOLD, Albert-GOLD. Each unique victory circle has actually been counted 6 times! Thus we must divide 24 by 6 to find the number of unique victory circles. There are actually only unique victory circles that contain 3 GOLD medalists. FINALLY, then, we have the following: 
(Combination 1) 24 unique GOLD-SILVER-BRONZE victory circles. (Combination 2) 12 unique GOLDGOLD-SILVER victory circles. (Combination 3) 12 unique GOLD-SILVER-SILVER victory circles. (Combination 4) 4 unique GOLD-GOLD-GOLD victory circles. Thus, there are unique victory circles.

This question is not as complicated as it may initially seem. The trick is to recognize a recurring pattern in the assignment of the guards. First, we have five guards (let's call them a, b, c, d, and e) and we have to break them down into pairs. So how many pairs are possible in a group of five dist inct ent ities? We could use the combinations formula: nCr. where n is the number of items you are selecting from (the pool) and k is the number of items you are selecting (the subgroup). Here we would get 5C2 = 10. So there are 10 different pairs in a group of 5 individuals. Ho wever, in this particular case, it is actually more helpful to write them out (since there are only 5 guards and 10 pairs, it is not so onerous): ab, ac, ad, ae, bc, bd, be, cd, ce, de. Now, on the first night (Monday), any one of the ten pairs may be assigned, since no one has worked yet. Let's say that pair ab is assigned to work the first night. That means no pair containing either a or b may be assigned on Tuesday night. That rules out 7 of the 10 pairs, leaving only cd, ce, and de available for assignment. If, say, cd were assigned on Tuesday, then on Wednesday no pair containing either c or d could be assigned. This leaves only 3 pairs available for Wednesday: ab, ae, and be. At this point the savvy test taker will realize that on any given night after the first, including Saturday, only 3 pairs will be available for assignment. Those test takers who are really on the ball may have realized right away that the assignment of any two guards on any night necessarily rules out 7 of the 10 pairs for the next night, leaving only 3 pairs available on all nights after Monday. The correct answer is Choice D; 3 different pairs will be available to patrol the grounds on Saturday night.

The key to this problem is to avoid list ing all the possibilit ies. Instead, think of an arrangement of five donuts and two dividers. The placement of the dividers determines which man is allotted which donuts, as pictured below: 

In this example, the first man receives one donut, the second man receives three donuts, and the third man receives one donut. Remember that it is possible for either one or two of the men to be allotted no donuts at all. This situation would be modeled with the arrangement below: 

Here, the second man receives no donuts. Now all that remains is to calculate the number of ways in which the donuts and dividers can be arranged: There are 7 objects. The number of ways in which 7 objects can be arranged can be computed by taking 7!:

However, the two dividers are identical, and the five donuts are identical. Therefore, we must divide 7! by 2! and by 5!:

48356 + 5412 – 736 = x * 40
53032 = x * 40
x = 1325.8

Answer – B. 20 L Explanation : 20/15 = x/x-5 x = 20

Change, HAVE → HAS,
Verb → Singular form.
MANY A is always followed by a singular noun and singular verb.

Change, in which there have been → in which there has been.

Bode: refers to forecasting/indicate/augur

Something bodes well when the clues or signs imply a favorable outcome.

Ex: Because I am a light sleeper, the dog's barking does not bode well for my desire to take a nap.

Definite  and certain are the opposite of indicate and incorrect.

Clues is irrelevant.

Only augur meaning foretell/portend/signify/bode is correct.

Hence, option D is correct.

'has always been' and 'further' control the words. The word 'but' does not warrant a contrast in this sentence because it has the sense of 'contrary to expectations', or simply, 'and' can easily replace 'but'. So the sentence communicates a 'bad to worse' situation. Hence option 3.

After looking at the given parts  P, Q, R, and S, it can be concluded that the correct order will be:  

When he → heard the hue and cry at midnight → he was nervous and → did not know → what to do. 

So, correct order is RQPS.

The weather outside was extremely pleasant and hence we decided to enjoy a morning ride in the open.