**Elitmus Syllabus ****| Syllabus for Elitmus pH Test**

### Elitmus ph Test Quantitative Aptitude Question Paper

**Question 1.** Find the number of ways you can fill a 3×3 grid (with 4 corners defined as a,b,c,d) if you have 3 white marbles and 6 black marbles

- 9 C 3
- 6 C 3
- 9 C 3 + 6 C 3
- (9 C 3) + (6 C 3)/3!

**Solution:**

- As the marbles are identical the required answer is 9C3 or 9C6 i.e we can start filing the grid either with white or black marbles.

We have 3 identical white marbles which can be placed in 9 blocks in 9C3 ways and the rest 6 blocks can be filled with 6 white marbles so the required ways of filling the grid is 9C3.

Even if we start filling the 3×3 grid with black marbles the required permutation is 9C6.

ANS. A) 9C3

**Question 2.** How many values of C in the equation x^{2} – 5x + c result in rational roots which are integers?

- 1
- 3
- 6
- Infinite

**Solution:**

The value of x can be calculated as:

For c = 0 , 4 , 6 , -6, -14, -24 , -36, -50 ……….and so on we can get rational roots which are

integers.

Hence the Answer is **D: Infinite.**

**Question 3.** If 1/a + 1/b + 1/c = 1/(a+b+c) where a + b + c != 0 , abc != 0, what is the value of (a+b)(b+c)(c+a)?

- Equals 0
- Greater than 0
- Less than 0
- Cannot be determined

**Question 4**. A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can this natural number have?

- Either 1 or 2
- Either 1 or 3
- Either 2 or 3
- Either 1,2 or 3

**Solution:**

**Ans) Either 1 or 2**

Check on 2^{9} i.e 512 , 3^{9}, 5^{9 }which have only 1 prime factor and 80 , 48 which are having 2

prime factors and total of 10 divisors.

Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a

number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have

minimum 8 such factors with xyz as the nos or the factor of the nos.

When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that

is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then

we will get at least 12 divisors. So we don’t get 3 prime factors with 10 divisors.

**Question 5**. Eden park is a cricket field while Jubilee park is a football field in Mastnagar .In Mastnagar, all cricket fields are circular and football fields are rectangular or square. Along the boundary of all fields there are advertisement displays. In Mastnagar, the length if advertisement displays has to be the same across all the playing fields. Area of Jubilee park is 468 sqm. The farthest distance between between any two points in Jubilee park football field is 10√10 m. Find the appropriate area of Eden park?

- 1936 m
^{2} - 616 m
^{2} - 281 m
^{2} - Inadequate Data

**Solution:**

Let us consider the football field to be rectangular with dimensions L X B. Even if the

football field is square then we will get L = B.

Given Area of jubilee park = 468sqm.

Area of rect. park = LxB = 468** …… (i)**

Farthest distance of field is diagonal ;

Therefore length of diagonal ; D = √L^{2}+B^{2 }= 10√10

or L^{2}+B^{2} = 1000** …….. (ii)**

Therefore; (L + B)^{2}= L^{2}+B^{2} + 2LB = 1000 + 2 x 468 = 1936

or , (L+B) = √1936 = 44

Therefore, Perimeter of rect. Field = 2 (L + B) = 2 x 44 = 88m

a/c to question; Perimeter of rectangular field = Circumference of Circular field

therefore, Circumference of Circular field = 2πR = 88m

Radius of circular field = 14m

Therefore Area of circular field = π142 = 616 sqm

**ANS. B) 616 m ^{2}**

**NOTE:** Questions on mensurations are not very tough but require some typical calculations

and logic. Proceed step by step and you will get the answer

**Question 6**. If v,w,x,y and z are non – negative integers , each less than 11, then how many distinct combinations of (v,w,x,y.z) satisfy v(11^4)+ w(11^3) + x(11^2) + y(11) + z = 151001

- 0
- 1
- 2
- 3

**Solution:**

** 11 ^{4}v+ 11^{3}w + 11^{2}x +11y +z = 151001 **

Taking 11 common from LHS seperating z and breaking 151001 as a factor of 11 with some remainder in RHS

11(11^{3}v+11^{2}w+11x+y) + z = 11×13727 + 4 [hence; z = 4]

Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;

11(11^{2}v + 11w +x) + y =11×1247 + 10 [y = 10]

11(11v + w ) + x = 11×113 + 4 [x = 4]

11v + w = 11×10 + 3 [v=10 , w=3]

So we get the unique solution for the above equaton.

**ANS. B) 1**

**Question 7**. The circle O having a diameter of 2cm has a square inscribed in it. Each side of the square inscribed in it. Each side of the square is the then taken as a diameter to form 4 smaller circles O”. Find the total area of all four O” circles which is outside the circle ).

- 2
- pi – 2
- 2- pi/4
- 2- pi/2

**Solution: **

AO = OB = 1cm (radii of circle O)

Area of quarter part of circle O = (1/4) pi 1^{2} = pi/4

AB is the side of square inscribed in circle O hence AB act as a diameter to the circle O”

AB = √2 (diameter of Circle O”)

Area of circle O’’ = pi (1/√2)^{2} = pi /2

Half Area of Circle O’’ = pi/4

Area of triangle (AOB) inside circle O” =1/2 x 1 x1 = 1/2

Area of two sectors (P and Q) of circle O”

= Half Area of Circle O’’- Area of triangle AOB

= (pi/4) – 1/2

Required Area = Area of circle O” – (Quarter area of circle O + area of sectors P & Q)

= pi/2 – (pi/4 + pi/4 – 1/2 ) =1/2

Therefore overall area for 4 such circles = 4 x 1/2= 2

**ANS. A) 2**

**NOTE:** Though questions based on geometry are time consuming but u will definitely get

the answer if u r good in geometry else leave such questions and try to solve other

questions. One geometrical figure inscribed in another geometrical figure are the questions

they commonly ask.

**Question 8**. How many six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place?

- 402
- 528
- 648
- 720

**Solution:**

Take the case of each digits separately and find the possible combinations.

1 |

2 | 5! = 120 |

3 | 5! = 120 |

2 or 6 | 4 | 4! x 2 = 48 |

5 | 5! = 120 |

6 | 5! = 120 |

Therefore, total 648.

ANS: C) 648

**Question 9**. Amit can complete a piece of work in 2.25 days. Badri takes double the time taken by Amit. Chetan takes double that of Badri, and Das takes double that of Chetan to complete the same task. They are split into two groups (of one or more persons) such that the difference between the times taken by the two groups (of one or more persons) such that the difference between the times taken by the two groups to complete the same work is minimum. What could be the composition of the faster group?

- Amit and Das
- Badri and Chetan
- Badri, Chetan and Das
- Amit alone

**Solution:**

Badri, Chetan and Das can do the work in 4.5 , 9 and 18 days

They together can perform the work in 1/(2/9+1/9+1/18)

= 18/7 days i.e 2.57

Amit alone can do the job in 2.25 days. So this is the minimum difference between any

two combinations.

**ANS. D) Amit alone**

**Question 10**. PR is tangent to a circle at point P. Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and PR = 6 units , then find the ratio of the perimeter of triangle PMR to triangle PQR.

- 11/20
- 3/5
- 13/20
- 18/25

**Solution:**

Given PQ = 8 cm (Diameter of Circle)

PR = 6 cm (Tangent to the circle)

Therefore QR = 10cm

Let QM = x and PM = y ; MR = 10-x

PM^{2} + MQ^{2} = PQ^{2} or x^{2} + y^{2} = 64 …….. (i)

MR^{2} + PM^{2 }= PR^{2}

or (10 – x)^{2 }+ y^{2} = 36 …… (ii)

Subtracting (ii) from (i)

X^{2} – (10-x)^{2}= 28

(x + 10 –x )(x – 10 +x) = 28

10(2x -10) = 28

x = 6.4

By solving we get y = 4.8

Perimeter of PMR / Perimeter of PQR= (4.8+3.6+6) /(8+10+6)

= 14.4/24

= 3/5

**ANS. B) 3/5**

**Question 11**. Heinz produces tomato puree by boiling tomato juice. The tomato puree has only 20% water while the tomato juice has 90% water. How many litres of tomato puree will be obtained from 20 litres of tomato juice?

- 2 litres
- 2.4 litres
- 2.5 litres
- 6 litres

**Solution:**

20 lit of tomato juice contain 2 lit of tomato (rest 18lit i.e 90% are water)

Tomato puree has 80% of tomato = 2 lit

Total Tomato puree = 2 x 100 /80 = 2.5 lit

**ANS. C) 2.5 L**

**Question 12**. Abhishek and Aishwarya pick up a ball at random from a bag containing M red and N yellow colored balls, one after the other, replacing the ball everytime till one of them gets a red ball. The first one to get a red ball is declared the winner. Of Abhishek begins the game and the odds in favor of his winning the game are 3 and 2, then find the ration M,N.

- 1:1
- 1:2
- 2:3
- 3:2

**Question 13**. A fresher recruitment event of hire all samart people ltd(HASPL) at elitmus happens in 2 cities Banglore and delhi. The Interview call is sent to everyone who is above 70th percentile(top 30% of the pool)70% and 80% of called people accept the interview call in Banglore and Delhi respectively. 90% and 80% of the people who accepted the call,turned out on the day of interview respectively in Banglore and Delhi.The ‘offer’ ratio is 2 out of 3 and 3 out of 4 of the people who turned up in Banglore and delhi respectively. If amongst people who apply there are 1000 people above 70th percentile in each of the locations, what percentage of 70th percentile from elitmus got offered in HASPL when results of both location are taken together?

1)60%

2)50%

3)45%

4)41%

**Solution: **

Let x people are above 70 percentile in Bangalore and y people are above 70 percentile

in Delhi who got a call for interview.

People who accept the call in Bangalore = 70% of x.

Those who attended the interview in Bangalore = 90% of 70% of x i.e 63% of x.

Those who got offered in Bangalore = 2/3 of 63% of x i.e 42% of x.

People who accept the call in Delhi = 80% of y.

Those who attended the interview in Delhi = 80% of 80% of y i.e 64% of y.

Those who got offered in Bangalore = 3/4 of 64% of y i.e 48% of y.

Given; x=1000 and y=1000

Those who were offered job = 42% of 1000 + 48% of 1000 = 900

i.e 900 got offered out of 2000(70+ percentile) which is 45% of the total called for the

interview.

**ANS. 3) 45%**

**Question 14**. What is the value of log_{e}(e(e(e….)^{1/2})^{1/2})^{1/2} ?

- 0
- 1/3
- 1/2
- 1

**Solution:**

log_{e}(e(e(e….)^{1/2})^{1/2})^{1/2}

(e(e(e….)^{1/2})^{1/2})^{1/2}= e^(1/2 + 1/4 +1/8 +….) = e^1

Log(e^1) = log e = 1

**ANS. D) 1**

**Question 15**. For Rs. 600 Mr. Karan can buy a maximum of 24 cups of tea and for Rs 1000 he can buy a maximum of 15cups of Coffee. He has Rs 2200 in his pocket and purchases 8 cups of tea. What is the maximum number of cups of coffee Mr. Karan can purchase with the remaining money?

- 33
- 32
- 31
- 27

**Solution:**

For Rs. 600 Mr. Karan can buy a maximum of 24 cups of tea (some money may be left)

Therefore for 8 cups of tea he requires a maximum Rs 200 (Even less then 200)

For remaining 2000(or more) he can buy atleast 30 cups of coffee and may be an extra

cup of coffee with the left money.

So, the answer can be 31.

**ANS. C) 31**

**Question 16.** The square of a two digit number is divided by half the number. After 36 is added to the quotient, this sum is then divided by 2. The digits of the resulting number are the same as those in the original number, but they are in reverse order. How many numbers fulfill the above criteria?

- 2
- 4
- 6
- 7

**Solution:**

When the square of the nos is divided by half of the nos the nos is twice of itself.

After adding 36 to it and then divided by 2 the nos increases by 18.Let xy be the nos (x be the tens digit and y unit digit)

Therefore; 10x + y + 18 = 10y + x (From question)

or 9x + 18 =9y

or x + 2 = y

Putting x = 1 to 7 we get the required nos.

ANS. D) 7

**Question 17**. In an opinion poll 78% of those asked were in the favour of atleast one of the three sportspersons to be included in the commonwealth organizing commitee Saina, Dhoni and Anand. 50% of those asked favoured Saina, 30% favored Dhoni 20% favored Anand.If 5% of those asked favoured all of them,what percentage of those asked favoured atleast one sportsperson ?

A)61

B)56

C)51

D)22

**Question 18**. A basket ball is dropped from a height of 20 feet.It bounces back each time to a height which is one half of hthe height of the last bounce. How far approximately will the ball have travelled before it comes to rest?

- 30 feet
- 40 feet
- 60 feet
- Can not be determined

**Solution:**

This is the simplest question (Simple G.P problem). The questions is : How much did the ball travel. Here is the catch.

When it was dropped from 20 feet. It traveled** 20 feet** to reach the ground then bounced 10 feet and then came back 10 feet i.e. total **20 feet** in this bounce, then came 5 feet + 5 feet = 10 feet to again reach ground.

Sum to infinity of a G.P is a/(1-r) .

Here a= 20 and r = 1/2

(20 + 10 + 10 + 5 + 5 + 2.5 + 2.5 + 1.25 + 1.25 + ………………)

= 20 + (20 + 10 + 5 + 2.5 + ….) = 20 + 20/1-(1/2) = 20 + 20 / (1/2) = 20 +40 = 60

ANS. C) 60 feet

**Question 19**. PT usha and shelly John decided to run a marathon between Ramnagar and jamnagar. Both start from Ramnagar at 1pm. on the way there are 2 town Ramghar and Rampur separated by a distance of 15 km. pt usha reaches Ramghar in 90 minutes running at a constent speed of 40 km/hr. She takes another 30 minutes to reach Rampur. Between Rampur and Jamnagar she maintain an average speed of V km/hr(Where V is whole number). Shelly John being a professional marathon runner maintain a speed of 18 km/hr. they both reaches jamnagar at ‘N’ hour (Where N is a whole number). What could be the total time taken by PT Usha?

options

A)5 hours

B)15 hours

C)41 hours

- D) All of above

Solution:

P.T.Usha took 90 min to reach Ramnagar with a speed of 40 km/hr

i.e. P.T.Usha travels 60 km (distance between Ramnagar and Ramgarh)

Time taken by her from Ramgarh to Rampur = 30 min

From Rampur to Jamnagar she maintains a constant speed of V km/hr

(where V is a whole nos)

Let N be the total time taken by P.T.Usha and Shelly John to travel from Ramnagar to

Jamnagar

(where N is a whole nos)

Speed of Shelly John = 18 km/hr

Total distance travelled by Shelly John in N hr = 18N …….(i)

P.T.Usha took 2 hr to reach from Ramnagar to Rampur.

She took (N-2) hr to travel from Rampur to Jamnagar with a speed of V km/hrTherefore distance between Rampur to Jamnagar = (N-2)V km

Total distance = 60km + 15km + (N-2)V km = 75 + (N-2)V ….(ii)

Equating both equations;

18N = 75 + (N-2)V

or V = 18N -75 / (N – 2)

- a) For N = 5 ; we get V = 5 (a whole nos)
- b) For N = 15; we get V = 13 (a whole nos)
- c) For N = 41; we get V = 17 (a whole nos)

**ANS. D) All of above**

**Question 20**. In a certain examination paper, there are n questions for j=1.2…n. There are 2^n-1 students who answered j or more questions wrongly. If the total numebr if wrong answers is 4095, then the value od n is?

- 12
- 11
- 10
- 9

Elitmus is intelligent enough to find out that the paper has leaked and so they might change the structure a little, So be Cautious!!