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Most Repeated Question in Elitmus previous exams - Aptitude Section

Question 1] For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is
A. n/2
B. n-1
C. n
D. None of these

Explanation :

Let A = { a1, a2, a3, a4, ................, an }
Initially, the function 'h' is applied to a1 and a2 and obtain a G.C.F.

Now, the function 'h' is applied to the HCF of a1 and a2 and the next number a3. This process is continued till the last number.

The final G.C.F is the G.C.F of the set A and the number of times we used the function 'h' is 1 [for a1 & a2 ] + (n-2) [for the rest of the values] i.e n-1.

Question 2] How many three digit number satisfies the following condition ?

a) When dividd by 11 or 12 leave remainder 7& 8 Respectively.

b) When divided by 33 or 24 lweave remiander 29 & 20 resp.

c) When divided by 7 or 8 leave remainder 4 in each case.
A.     5
B.     3
C.     1
D.     None of these

Explanation :

Number Satisfying  the second condition also satisfies the first condition.

They are of the form

N = 33p+29& N = 24q+20.

N+4 = K LCM of (33, 24)

N+4 = 264k - 4

N = 260 ( When K = 1) or 524 ( When K = 2) or 788 ( When K = 3).

And, only 788 Satisfies the third condition also.

Question 3 ] Let n! = 1 x 2 x 3 ........n for integer n ≥ 1. If p = 1! + (2 x 2!) + + (3 x 3!) + .......+ (10 x 10!), then p + 2 when divided by 11! leaves a remainder of
A.     10
B.     0
C.     7
D.     1

Explanation :
P = 1 + 2.2! + 3.3!+ ….10.10!.
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10!.
= 2! – 1! + 3! – 2! + ….. 11! –10!.
= 1 + 11!.
Hence, the remainder is 1.

Question 4] What is the remainder when N= (1! +2! +3! +4! +... + 1000!)40 is divided by 10?
A.     1
B.     2
C.     3
D.     7

Explanation :

We have to find out remainder of (1! + 2! + 3! ... 1000!)40 from 10.
=> We have to find out last digit of (1! + 2! + 3! ... 1000!)40

Last digit of n! where n > 4 will be 0 and will have no impact on the answer.
=> We have to find out last digit of (1! + 2! + 3! + 4!)40.

Last digit of (1! + 2! + 3! + 4!)40
= Last digit of (1 + 2 + 6 + 24)40
= Last digit of 3340
= Last digit of 340
= Last digit of 8110
= Last digit of 110
= 1.

Question 5] A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminum of width 2 units, as shown below. The hole is punched such that circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

The proportion of the sheet area that remains after punching is :-
A.     (π + 2)/8
B.     (6 - π)/8
C.     (4 - π)/4
D.     (π - 2)/4
E.     (14 - 3π)/6

Explanation :

First, we have to expand the diagram to the following :-

Let PQRS be the square. Let A&B be the points of intersection of the circle on the square. Join AB. It is said that the diameter of the circle is along the same line as that of the diagonal of the square. So draw the diagonal PR of the square. Let it meet the circle at C. So PC is the diameter of the circle.

We know, ∟APB = 900 (PQRS is a square). Also, angle formed by a semicircle is 900. Hence the sector AB is semicircle, and AB is diameter.

Let the meeting point of AB and PC be O, both diameters meet at the centre of the circle, and hence O is the centre.

Now, since PC is the diameter of the circle, ∟PAC and ∟PBC is 900, and since AB is a diameter, ∟ACB is 900.

Now, in ACBP, we know all the angles are 900, and CP = AB (they are diameters of a circle). BP and AC are diagonals of the quadrilateral ACBP. Hence ACBP is a square.

Diagonals are perpendicular bisectors in a square and hence ∟POB = ∟BOC = ∟COA = ∟AOP = 900, and PO = CO = AO = BO = 1.

Now, the portion of the circle outside the square is :

Minor sector AP + Minor sector BP – Area of ∆AOP – Area of ∆BOP.

Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)

Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)

Area of ∆AOP = ½*b*h = ½*1*1 = ½.

Area of ∆BOP = ½*b*h = ½*1*1 = 1/2.

So, the portion outside the circle is (π/2) – 1 or (π – 2)/2.

We know the whole circle area is π*r2, which is π here. (as r = 1).

So, the area inside the square is π – ((π – 2)/2) = (π + 2)/2.

Initial area of square = 2*2 = 4.

Area of square remaining = 4 – ((π + 2)/2) = (6 - π)/2.

Proportion of the square remaining = ((6 - π)/2)/4 (Change in area / Original area) i.e (6 - π)/8.

Question 6] Two sides of a plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles.

What is the area of the plot?
A.     768 sq. m
B.     534 sq. m
C.     696.5 sq. m
D.     684 sq. m

Explanation :

CE = sqrt (252−202) = 15.

(Since DBC is isosceles triangle.)

Assume ABCD is a quadrilateral where AB = 32 m, AD = 24 m, DC = 25 m, CB = 25 m and ∠DAB is right angle.

Then DB = 40 m because ∆ADB is a right-angled triangle and DBC is an isosceles triangle.

So area of ∆ ADB = (1/2) x 32 x 24 = 384 sq. m.

So area of ∆ BCD = 2 x (1/2) x 15 x 20 = 300 sq. m.

Hence, area of ABCD = 384 + 300 = 684 sq. m.

Question 7 ] Two circles with centers P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees?
A.     Between 0 and 90
B.     Between 0 and 30
C.     Between 0 and 60
D.     Between 0 and 75
E.     Between 0 and 45

Explanation :

It says P & Q falls outside the intersection area of the circle.

Let us draw the diagram first.

The maximum point where P & Q can be in the circle is represented by the 2 small dashes P1 and Q1, when they can be on the circumference of the other circle.

Suppose P & Q are on the circumference of the other circle. (Represented by P1 and Q1)

Then AP1 = r1 (Radius of 1st circle)

AQ1 = r2 (Radius of 2nd circle)

Also P1Q1 = r1 = r2 (As radius of both the circles are same)

So AP1 = AQ1 = P1Q1.

Hence ∆AP1Q1 is equilateral, and ∠AQ1P1 = 600.

Suppose P1 and Q1 drift away from the circle, and are at the points P & Q represented in the diagram.

Then the value is definitely less than 600.

If P and Q are col-linear with A, then the angle is 00.

So, the maximum value is 600, and minimum 00

Question 8] A and B play a series of 5 games in badminton. A's chance of winning s single game against B is 2/3. Find the chance that A wins exactly 3 games.
A.     64/81
B.     35/128
C.     80/243
D.     None of these

Explanation :

Here, each game is a trial and A's winning the game is a success. All games are independent , so, using the binomial distribution.

P(X) = nCx px(1 - p)n-x.

Where, x= wins exactly 3 games

=> x = 3.

p= 2/3, 1-p=1/3, n=5.

P(3) = 5C3 (2/3)3(1/3)2 .

=> 10 x (23/35).

=> 80/243.

Question 9 ] A packet of 10 electronic components is known to include 3 defectives. If 4 components are randomly chosen and tested, what is the probability of finding among them not more than one defective component.
A.     0.6517
B.     0.868
C.     0.4483
D.     None of these

Explanation :

Here, the trials are independent , since the probability of defective component in any trial is not affected by the occurrence or non-occurrence of a defective component in any other trial. So, it is a case of binomial distribution.

No. of trials = n = 4.

Occurrence of a defective component is a success.

Then, the probability of success = p = 3/10.

The probability of failure = 1- p = 7/10.

Components not more than one defective one means x ≤ 1 i.e x = 0.

Using the  formula,

P(x) = ncxPx(1-P)n-x.

Since,

=> P(x ≤ 1)  = P(x = 0) or P(x = 1)

= P(0) + P(1) .

= 4c0(3/10)0 x (7/10)4  +  4c1(3/10)1x (7/10)3 .

= (7/10)4 + 4 x (3/10) x (7/10)3.

= 0.6517.

Hence, the required probability is 0.6517.

Question 10] There are 10 white, 8 red and 6 green balls on a billiard board. Find the number of ways in which one or more balls can be put in the pocket.
A.     672
B.     692
C.     693
D.     None of these

Explanation :

Here, number of balls to be put is not given, but the minimum number i.e 1 and the maximum number of balls (10 + 8 + 6 ) i.e 24 is known.

At least one ball is to be included in the required selection.

Hence, the required number of ways is :-

=> Selection without any restriction - none.

=> (10 + 1) x (8 + 1) x (6 + 1) - 1. [ Since,there is only one way when none is selected ]

=> 11 x 9 x 7 - 1.

=> 692.

Question 11] There are 5 different green dyes, 4 different blue dyes and 3 different red dyes. Find the number of combinations of dyes that can be chosen taking at least one green and one blue dyes.
A.     465
B.     3720
C.     4096
D.     None of these

Explanation :

Here, the dyes of same color are also different. But the selection is to be made such that at least one green and one blue dye are included.

Hence, the required number of ways of selection is (2n1-1)(2n2-1)(2n3),

where, n1 = 5 different green dyes (at least one to be included)

n2 = 4 different blue dyes (at least one to be included)

n3 = 3 different red dyes (zero or more to be included)

=>  (25-1)(24-1)(23).

=> 31 x 15 x 8.

=> 3720.

Question 12] In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 III are not defective. What is the percentage of defective in the stockpile?
A.     3%
B.     5%
C.     2.5%
D.     4%

Explanation :

Let there be 100 products in the stockpile. 40% and 30% were manufactured by M1 and M2 respectively. So, 30% are manufactured by M3.

Products from M1 = 40, from M2 = 30 and from M3 = 30.

Number of defective products from M1 = 0.03 x 40 = 1.2, from M2 = 0.01 x 30 = 0.3 and from M3 = 0.05 x 30 = 1.5

Therefore, total number of defective products = 3.

Question 13] A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than
A. 6
B. 12
C. 3
D. 9

Explanation :

Let the number of correct answers be 'x', number fo wrong answers be 'y' and number of questions not attempted be 'z'.

Thus, x + y + z = 50.                                                      ................(i)

And, x - (y/6) - (z/6) = 32.                                           ................(ii)

The 2nd equation can be written as

=> 6x -y - z = 192.                                                          ..................(iii)

Adding equations (i) and (iii), we get :-

=> 7x - y = 242.

=> x = (242/7) + y.

Since, x and y are both integers, y can't be 1 or 2.

The minimum value that y can have is 3.

Question 14] At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes and 1 order of fries for exactly Rs 120. At the same place it would cost Rs 164.5 for 4 burgers, 10 shakes and 1 order of fries. How much would it cost for an ordinary meal of 1burgers, 1 shakes and 1 order of fries ?
A.     Rs. 31
B.     Rs 41
C.     Rs 21
D.     Cannot be determined

Explanation :

Let the cost of 1 burger, 1 shake and 1 fries be x, y and z.
Then,
3x + 7y + z = 120                              ... (i)
4x + 10y + z = 164.5                        ... (ii)

After subtracting (i) from (ii) :-
x + 3y = 44.5                                      ... (iii)

Multiplying (iii) by 4 and subtracting (ii) from it, we find :-
2y – z = 13.5                                      ...(iv)

Subtracting (iv) from (iii), we get-

x + y + z = 31.

Question 15] For a Fibonacci sequence, from the third term onward, each term in a sequence is the sum of the previous two terms in that sequence. If the difference of squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?
A.     147
B.     76
C.     123
D.     can't be determined

Explanation :

Let the 6th and the 7th terms be x and y.
Then 8th term = x + y
Also y2 – x2  = 517
=> (y + x)(y – x) = 517 = 47 × 11 .
So, y + x = 47.
y – x = 11.
Taking y = 29 and x = 18,

we have 8th term = 47,

9th term = 47 + 29 = 76 and

10th term = 76 + 47 = 123.

Question 16] Consider the following series 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4 …. and so on. What is the sum of the first 1000 terms of the series?
A.     21088
B.     21120
C.     20118
D.     21808

Explanation :

1+1+2+2+2+2+3+3+3+3+3+3+4+4+.... upto 1000 terms

= 1*2 + 2*4 +3*6 .....+ 31*62 + 32 *8

= 2(1^2 + 2^2 + 3^2 ....+ 31^2) + 32*8

= 2*31(31 +1)(2*31 + 1)/6 + 32*8, as 1^2 + 2^2 + ... n^2 = n(n+1)(2n+1)/6

= 21088.

Question 17] If (1/3) log3M + 3log3N = 1+ log0.0085, then
A.     M^9 = 9/N
B.     N^9 = 9/M
C.     M^3 = 3/N
D.     N^9 = 3/M

Explanation :

=> (1/3) log3M + 3log3N = 1+ log0.0085

=> (1/3) log3M1/3N = 1 + (log10 - log2) / (log8 - log1000).

=> (1/3)log3M1/3N = 1 - (1 - log2) / 3(1 - log2).

=> (1/3) log3M1/3N= 1 - 1/3 = 2/3.

or, log3M1/3 = 32/3.

or, MN9 = 32.

or N9 = 9/M.

Question 18] Two typists undertake to do a job. The second typist begins working one hour after the first. Three hours after the first typist has begun working, there is still 9/20 of the work to be done. When the assignment is completed, it turns out that each typist has done half the work. How many hours would it take each one to do the whole job individually ?
A.     12 hr and 8 hr
B.     8 hr and 5.6 hr
C.     10 hr and 8 hr
D.     5 hr and 4 hr

Explanation :

Let the first typist takes X hours and the second Y hours to do the whole job.

When the first was busy typing for 3 hr, the second was busy only for 2 hr. Both of them did 11/20 of the whole work.
=> 3/X + 2/Y = 11/20.
When the assignment was completed, it turned out that each typist had done half the work. Hence, the first spent X/2 hr, and the second, Y/2 hr.
And since the first had begun one hour before the second,

we have X/2 – Y/2 =1

i.e X -Y =2;

Hence, X = 10 hr, Y = 8 hr.

Question 19] One man can do as much work in one day as a woman can do in 2 days. A child does one third the work in a day as a woman. If an estate-owner hires 39 pairs of hands, men, women and children in the ratio 6 : 5 : 2 and pays them in all Rs. 1113 at the end of the days work. What must the daily wages of a child be, if the wages are proportional to the amount of work done?
A.     Rs. 14
B.     Rs. 5
C.     Rs 20
D.     Rs 7

Explanation :
Let work done by a man in one day = x
Then,
work done by a woman in 1 day = x/2.
work done by a child in 1 day = x/6.
estate owner hires total 39 persons
men : women : children = 6:5:2
Number of men = 39 × 6 / 13 = 18
Number of women = 39 × 5 / 13 = 15.
Number of children = 39 × 2 / 13 = 6.
Work done by 18 men in 1 day = 18x
Work done by 15 women in 1 day = 15x/2
Work done by 6 children in 1 day = 6x/6 = x.
Total amount given to 6 children = 1113 × x / ( 18x + 15x/2 + x) = 2226 / (36 + 15 + 2 ) = 42.
Daily wage of a child = 42 / 6 = Rs. 7.

Question 20] A person was fined for exceeding the speed limit by 10 mph. Another person was also fined for exceeding the same speed limit by twice the same. If the second person was traveling at a speed of 35 mph, find the speed limit.
A.     10 mph
B.     15 mph
C.     10 mph
D.     20 mph

Explanation :

Let ‘x’ be the speed limit. Person ‘A’ was fined for exceeding the speed limit by = 10mph

Person ‘B’ was fined for exceeding the speed limit by = twice of ‘A’ => 2*10mph=20mph

Given that the second person was traveling at the speed of 35mph => 35mph – 20mph = 15mph

Therefore the speed limit is =15 mph