Quant Test 2 Practice

N^th term of the series can be written as
tn=(4n+3)(4n+7)
=16n2+40n+21
Σtn=16Σn2+40Σn+21Σ1
=16n(n+1)(2n+1)6+40n(n+1)2+21n
here n = 23 (7, 11, 15….. 95 is an AP with common different 4 with 23 terms)
∑tn=16×23×24×476+20×23×24+21×23
= 80707

Let 'ab' be the two digit number. Where b ≠

0.

On interchanging the digits, the new number will be ‘ba’

As per the condition 10a+b > 3×(10b + a)

7a > 29b

For b = 1, a = {5, 6, 7, 8, 9}

For b = 2, a = {9}

For b = 3, no value of 'a' is possible.

Hence, there are a total of 6 such numbers

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
=> (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

Let 'x' and 'y' be the speed (in km/hr) of cars starting from both A and B respectively.
If they both move in east direction, then B will overtake A only if y > x.
Also, relative speed of both the cars when they move in east direction = (y - x) km/hr
It is mentioned that they take 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (y-x) km/hr.
Hence, (y - x) = 350/7 = 50 km/hr

Let the volume of the first and the second Solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture 

= (20 + 300S)

Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final Solution 

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = 12

= 50%

Hence, 50 is the correct answer.

Let the rate of filling of Type A and Type B pipes be a and b respectively.
Given 30×(10a + 45b) = 1 and 60×(8a + 18b) = 1
=> 30 × (10a + 45b) = 60 × (8a + 18b)
=> 10a + 45b = 16a + 36b
=> 3b = 2a or a = 1.5b
The total work = 30 × (10a + 45b) = 30 × (15b + 45b)
= 1800b
Required answer =1800b7a+27b=1800b10.5b+27b=48

Final quantity of alcohol in the mixture = 700700+175×(90100)2×[700+175]

= 567 ml

Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml

Hence, the percentage of water in the mixture = 308875×100

= 35.2 %

Among a group of n persons, number of matches played = n(n – 1)/2
Among the Junior participants, let the number of girls be n.
The number of matches played among girls
= n(n – 1)/2 = 153
=> n(n – 1) = 306 = 18 × 17 => n = 18
Number of boys = 43 – 18 = 25
The number of matches played between a boy and a girl = 25×18 = 450
Among the Senior level participants, let the number of boys be n.
The number of matches played between two boys
= n(n – 1)/2 = 276
=> n(n – 1) = 552 = 24 × 23 => n = 24
The number of girls = 51 – 24 = 27
The number of matches played between a boy and a girl = 27 × 24 = 648
Required answer = 450 + 648 = 1098

We want to maximize the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since a₅₂ is 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. a₂₇ be A.

(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

Since a₅₂ = a₂₇ + 25 and a₅₂ = 100

=> A = 100 – 25 = 75

a₂ + a₃ + … + a₅₂ = 75×51 = 3825

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

Hence a₁ = 3848 – 3825 = 23

P = {1,2,3,4} and Q = {2,3,5,6,}

PΔQ = {1, 4, 5, 6}

R = {1,3,7,8,9} and S = {2,4,9,10}

RΔS = {1, 2, 3, 4, 7, 8, 10}

(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}

Thus, there are 7 elements in (PΔQ)Δ(RΔS) .

hence, 7 is the correct answer.