## Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2), …a(base52)

Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), …, a(base52). If a(base52) = 100, then the largest possible value of a(base1) is?

a) 45

b) 20

c) 48

d) 23

Solution:

We want to maximize the value of a1, subject to the condition that

(a2+ a3 +⋯….+ a52 )/51 -(a1+ a2 +⋯….+ a52 )/52 = 1

Since a(base52) = 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3 ….a51. The only way to do this is to assume that a2, a3…. A52 are in an AP with a common difference of 1.

Let the average of a2, a3, …, a52 i.e.

(a2+ a3 +⋯….+ a52 )/51= a27 =A ( using the average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

So a52  = a27 + 25*1 = a27 + 25  and

given a52 = 100

=>  a27 = A = 100 – 25 = 75

a2 + a3 + … + a52 = 75×51 = 3825

Given a1 + a2 +… + a52  = 52(A – 1) = 3848

Hence a1 = 3848 – 3825 = 23

Ans : 23

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## Slot -2 – Quantitative Aptitude – Modern Maths – Set Theory – If A = {6^2n -35n -1: n = 1,2,3,…}

If A = {6^2n -35n -1: n = 1,2,3,…} and B = {35(n-1) : n = 1,2,3,…} then which of the following is true?

a) Neither every member of A is in B nor every member of B is in A

b) Every member of A is in B and at least one member of B is not in A

c) Every member of B is in A.

d) At least one member of A is not in B

Answer: b) Every member of A is in B and at least one member of B is not in A.

Solution:
Given, A = 36^n – 35n – 1 = 36^n – 1^n – 35n

Since a^n – b^n is divisible by a – b for all positive integral values of n, So A is a

multiple of 35 for any integral value of n and B is a set containing all the

multiple of 35 including 0.

Hence, every member of A is in B but not every element of B is in A.

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## Slot -2 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A water tank has inlets of two types A

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Suppose inlets of type A fill A liters per minute and type B fills B liters per minute.

So, capacity of tank = 30(10A + 45B) = 60(8A + 18B)

⇒10A + 45B = 16A + 36B

⇒6A = 9B

⇒A = 1.5B

⇒Capacity of Tank = 30(15B + 45B) = 30*60B = 1800 B

Time taken to fill the tank with 7A and 27B, which is 10.5B and 27B, which is 37.5B = 1800 B / 37.5 B = 48 minutes

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## Slot -2 – Quantitative Aptitude – Number System – The smallest integer n for which

The smallest integer n for which 4^n > 17^19 holds, is closest to?
a) 33

b) 35

c) 37

d) 39

Solution:
Given , 4^n > 17^19
Or  16^(n/2) > 17^19

Now from the given options, the only possible value of n is 39 as for other

values n/2 will be less than 19 and 16 < 17.

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## Slot -2 – Quantitative Aptitude – Arithmetic – Time Speed and Distance – Points A, P, Q and B lie on the same line

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is?

a) 2 : 7

b) 2 : 9

c) 1 : 4

d) 1 : 2

Solution:
Car 2 and car 3 meets at Q, so ratio of their speed = AQ : BQ = 200 : 100 = 2:1

Car 1 and car 3 meets at P, so ratio of their speed = AP : BP = 100 : 200 = 1 : 2

So required ratio of speed of car 1 and car 2 = 1 : 4

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## Slot -2 – Quantitative Aptitude – Arithmetic – Simple Interest and Compound Interest – Gopal borrows Rs. X from Ankit at 8%

Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y?

Solution:
Interest to be repaid to Ankit at the end of the year = 0.08X

Interest that Gopal would receive from Ishan in two cases are as given.

Case I: if he lends X + Y Interest received = (X + Y) × 0.1 = 0.1X + 0.1Y
Interest retained by Gopal after paying to Ankit = (0.1X + 0.1Y) – (0.08X) = 0.02X + 0.1Y
Given that Interest retained by Gopal is same as that accrued by Ankit
So (0.02X + 0.1Y) = 0.08X
=> Y = 0.6X

Case II: if he lends X + 2Y Interest received = (X + 2Y) × 0.1 = 0.1X + 0.2Y
Interest retained by Gopal after paying to Ankit = (0.1X + 0.2Y) – (0.08X) = 0.02X + 0.2Y

Given that interest retained by Gopal would increase by 150

=> (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150
0.1Y = 150

> Y = 1500 and X = 1500×0.6= 2500
Hence X + Y = 2500 + 1500 = 4000

Ans : 4000

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## Slot -2 – Quantitative Aptitude – Geometry – Circles – A chord of length 5 cm subtends an

A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is?

a) 6√2

b) 8

c) 4√2

d) 5√3

Solution:

In triangle ODA, OA=AD cosec 30=5

So if the angle AOB is 120 degree, then angle AOD will be 120/2=60

In the case length of chord = 2×OA Sin 60=5√3

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## Slot -2 – Quantitative Aptitude – Arithmetic – Ratios – The scores of Amal and Bimal in

The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is?
a) 8 : 5

b) 3 : 2

c) 4 : 3

d) 5 : 4

Solution: let their scores were 11x and 14x and it increase by n then

(11x+n)/(14x+n)=47/56

616x+56n=658x+47n

42x=9n

n = 42x/9

So Bimal’s new score = 14x + 42x/9 = 168x/9

So required ratio = (168x/9)/14x = 4∶ 3

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## Slot -2 – Quantitative Aptitude – Algebra – Functions – Let f(x)=max{5x, 52-2x^2}

Let f(x)=max{5x, 52-2x^2}, where x is any positive real number. Then the minimum possible value of f(x) is?

Solution:
For f(x) to be minimum , 5𝑥 = 52 − 2𝑥^2
2𝑥^2+5𝑥 −52 =0
𝑥−42𝑥+13=0
𝑆𝑜 𝑥 = 4
Thus minimum value of f(x) = 5x = 5*4 = 20

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## Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – The arithmetic mean of x, y

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is?

Solution:
Given, (x+y+z)/3=80

x+y+z=240——1)

And (x+y+z+u+v)/4=75

x+y+z+u+v =375——2)

From eq 1) & eq 2) u+v =135

And from question , u+v =(x+2y+z)/2

Or 270=x+2y+z

y = 30 and x+z = 210

as x ≥ z
so x will be minimum if x =z

minimum value of x =210/2 = 105

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