EPFO Assistant Prelims Mock Test - 8 - Bank Exams MCQ

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

As 24:60 = (2/5)

Similarly, (120/300) = (2/5)

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.

(Two persons are designated between both the persons, except option D)

From both I and II, we find that Rahul is (35 - 25) = 10 years older than his brother, who was born in 1964. So, Rahul was born in 1954.

From I, we conclude that H is the only daughter of M. But this does not indicate that M has no son. The information given in II is immaterial.

Sunita has a very busy schedule. This means that she is industrious. But still she finds time for rest. This means that she is an organised person. So, both I and II follow.

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.

Volume of water in the reservoir = Area of cross-section of pipe × Empty rate × Empty time
Let the empty time be t.

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 

Answer – d) 46kg Explanation : Suppose initially price per kg of rice is 100 then their expenditure is 5000.
Now their expenditure is only increased by only 20% i.e – 6000.
Increased price of rice = 130.
So new consumption = 6000/130 = 46

LCM (6, 8, 15, 30) – 4 = 120 – 4 = 116

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

Custom duty      :      Income tax
14      :      26
7      :      13

Excise duty + custom duty      :      other
41% + 14%      :      10%
55%      :      10%
11      :      2

Given:

Total faculties in school F = 1820

Calculation:

It's given that the total number of faculties (Male and Female) in school F is 1820 

Female faculties = Total faculties - Male faculties 

But according to the chart male percentage in school F = 18%

Total faculties in school F = 1820

Total male faculties in school F = 8400 × 18/100 = 1512

Female faculties in school F = 1820 - 1512 = 308

Male faculties in school B = 8400 × 20/100 = 1680

∴ Required percentage = (1680 - 308)/1680 × 100

⇒ 1372/1680 × 100 = 81.66%.

∴ Female faculties in school F are 81.66% less than male faculties in school B.

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.

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!:

There are two possibilit ies in this problem. Eit her Kim and Deborah will both get chocolate chip cookies or Kim and Deborah will both get oatmeal cookies. If Kim and Deborah both get chocolate chip cookies, then there are 3 oatmeal cookies and 2 chocolate chip cookies left for the remaining four children. There are 5!/3!2! = 10 ways for these 5 remaining cookies to be distributed--four of the cookies will go to the children, one to the dog. (There are 5! ways to arrange 5 objects but the three oatmeal cookies are identical so we divide by 3!, and the two chocolate chip cookies are identical so we divide by 2!.) If Kim and Deborah both get oatmeal cookies, there are 4 chocolate chip cookies and 1 oatmeal cookie left for the remaining four children. There are 5!/4! = 5 ways for these 5 remaining cookies to be distributed--four of the cookies will go to the children, one to the dog. (There are 5! ways to arrange 5 objects but the four chocolate chip cookies are identical so we divide by 4!.) Accounting for both possibilities, there are 10 + 5 = 15 ways for the cookies to be distributed.

3.25 * 2 = 6.5
6.5 * 3 = 19.5
19.5 * 4 = 78

Event definition is: 15 or 16 or 17 or 18. 15 can be got as: 5 and 5 and 5 (one way) Or 6 and 5 and 4 (Six ways) Or 6 and 6 and 3 (3 ways) Total 10 ways. 16 can be got as: 6 and 6 and 4 (3 ways) Or 6 and 5 and 5 (3 ways) Total 6 ways. 17 has 3 ways and 18 has 1 way of appearing. Thus, the required probability is:(10 + 6 + 3 + l)/216 = 20/216 = 5/54.

Help is a transitive verb. So, it should have an object. Here, TO doesn't require as help is transitive. Therefore, to should be replaced by The. Correct Sentence: The Committee has made significant changes in the rules which will help the banks.

The correct sentence is
The number of tourists is expected to rise seven to ten percent in the next ten years.

'food inflation', 'acuteness of', 'into political focus' etc. control the blanks. Something is brought into political focus only when it is unrelenting (option 4) or persistent (option 5). Food inflation does not directly result in scarcity or shortage but necessarily creates insecurity. Hence option 2. The first part of options 1, 3 and 5 are nonsensical.

Being unsure of itself controls the first blank—unsure makes hesitantly the best option for the blank. Between chary, circumspect and wary, wary (cautious) fits the best. The difference between chary and wary is that: circumspect suggests less fear and stresses the surveying of all possible consequences before acting or deciding. Wary emphasizes suspiciousness and alertness in watching for danger and cunning in escaping it keeps a wary eye. Chary implies a cautious reluctance to give, act, or speak freely.