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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeIf √[(xa)/(xb)] + a/x = √[(xb)/(xa)] + b/x, and given that b ≠ a, then the value of x is
Correct
Explanation:
√[(xa)/(xb)] + a/x = √[(xb)/(xa)] + b/x
√[(xa)/(xb)] – √[(xb)/(xa)] = b/x – a/x
√[(xa)(xa)/(xa)(xb)] – √[(xb)(xb)/(xa)(xb)] = (ba)/x
(xa)/√[(xa)(xb)] – (xb)/√[(xa)(xb)] = (ba)/x
(xa – x + b)/√[(xa)(xb)] = (ba)/x
So x = √[(xa)(xb)]
Square both sides
x^2 = (xa)(xb)
solve, ax + bx = ab
so x = ab/(a+b)Incorrect
Explanation:
√[(xa)/(xb)] + a/x = √[(xb)/(xa)] + b/x
√[(xa)/(xb)] – √[(xb)/(xa)] = b/x – a/x
√[(xa)(xa)/(xa)(xb)] – √[(xb)(xb)/(xa)(xb)] = (ba)/x
(xa)/√[(xa)(xb)] – (xb)/√[(xa)(xb)] = (ba)/x
(xa – x + b)/√[(xa)(xb)] = (ba)/x
So x = √[(xa)(xb)]
Square both sides
x^2 = (xa)(xb)
solve, ax + bx = ab
so x = ab/(a+b) 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeIf x = 2√(30)/(√10 + √15), then the value of (x + √3)/(x – √3) + (x + √2)/(x – √2)
Correct
Explanation:
x = 2√10*√3/(√10 + √15)
so x/√3 = 2√10/(√10 + √15)
use componendo and dividendo
so x + √3/x – √3 = (2√10 + √10 + √15)/(2√10 – √10 – √15)
(x + √3)/(x – √3) = (3√10 + √15)/(√10 – √15)
Now
x = 2√15*√2/(√10 + √15)
so x/√2 = 2√15/(√10 + √15)
use componendo and dividendo
so x + √2/x – √2 = (2√15 + √10 + √15)/(2√15 – √10 – √15)
(x + √2)/(x – √2) = (3√15 + √10)/(√15 – √10)
So(x + √3)/(x – √3) + (x + √2)/(x – √2)
= (3√10 + √15)/(√10 – √15) + (3√15 + √10)/(√15 – √10) = 2Incorrect
Explanation:
x = 2√10*√3/(√10 + √15)
so x/√3 = 2√10/(√10 + √15)
use componendo and dividendo
so x + √3/x – √3 = (2√10 + √10 + √15)/(2√10 – √10 – √15)
(x + √3)/(x – √3) = (3√10 + √15)/(√10 – √15)
Now
x = 2√15*√2/(√10 + √15)
so x/√2 = 2√15/(√10 + √15)
use componendo and dividendo
so x + √2/x – √2 = (2√15 + √10 + √15)/(2√15 – √10 – √15)
(x + √2)/(x – √2) = (3√15 + √10)/(√15 – √10)
So(x + √3)/(x – √3) + (x + √2)/(x – √2)
= (3√10 + √15)/(√10 – √15) + (3√15 + √10)/(√15 – √10) = 2 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA invested Rs 7000 and B invested Rs 14,000. After 5 months, A withdrew Rs 2000. At the end of year, if B got a share of Rs 18,000, what is A’s share?
Correct
Explanation:
7000*5 : 5000*7 : 14000*12
Solve, 5 : 12
So 12/17 * x = 18,000
So A’s share = 5/17 * x = 5/17 * 17/12 * 18000Incorrect
Explanation:
7000*5 : 5000*7 : 14000*12
Solve, 5 : 12
So 12/17 * x = 18,000
So A’s share = 5/17 * x = 5/17 * 17/12 * 18000 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeTwo articles are sold at same price. The first was sold at the profit of 20% and other at the loss of 10%. What is the profit/loss % in whole transaction?
Correct
Explanation:
Use formula
[100(2010) – 2*20*10] / [(100+20) + (10010)] = + 2 6/7%Incorrect
Explanation:
Use formula
[100(2010) – 2*20*10] / [(100+20) + (10010)] = + 2 6/7% 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeABCD is a trapezium in which AB = CD and AD is parallel to BC. If AD = 6cm, BC = 10cm, and area of trapezium ABCD is 64sq. cm, then length CD is
Correct
Explanation:
Area of trap = ½ * (AD+BC) * DF
64 = 1/2*(6+10) * DF
So DF = 8
FC = (106)/2 = 2
So CD = √(DF^2 +FC^2) = √(8^2 + 2^2) = 2√17Incorrect
Explanation:
Area of trap = ½ * (AD+BC) * DF
64 = 1/2*(6+10) * DF
So DF = 8
FC = (106)/2 = 2
So CD = √(DF^2 +FC^2) = √(8^2 + 2^2) = 2√17 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeIf xsin^{3}ɸ + ycos^{3}ɸ = sinɸcosɸ ≠ 0 and xsinɸ – ycosɸ = 0, then value of (x^{2} + y^{2}) =
Correct
Explanation:
xsin^{3}ɸ + ycos^{3}ɸ = sinɸcosɸ
xsinɸ . sin^{2}ɸ + ycosɸ . cos^{2}ɸ = sinɸcosɸ
now xsinɸ – ycosɸ = 0 so xsinɸ = ycosɸ
so
xsinɸ . sin^{2}ɸ + ycosɸ . cos^{2}ɸ = sinɸcosɸxsinɸ . sin^{2}ɸ + xsinɸ . cos^{2}ɸ = sinɸcosɸ
xsinɸ . (sin^{2}ɸ + cos^{2}ɸ) = sinɸcosɸ
so xsinɸ = sinɸcosɸ
so x = cosɸ
now since xsinɸ = ycosɸ
solve by putting x = cosɸ, we get y = sinɸ
so x^{2} + y^{2} = 1Incorrect
Explanation:
xsin^{3}ɸ + ycos^{3}ɸ = sinɸcosɸ
xsinɸ . sin^{2}ɸ + ycosɸ . cos^{2}ɸ = sinɸcosɸ
now xsinɸ – ycosɸ = 0 so xsinɸ = ycosɸ
so
xsinɸ . sin^{2}ɸ + ycosɸ . cos^{2}ɸ = sinɸcosɸxsinɸ . sin^{2}ɸ + xsinɸ . cos^{2}ɸ = sinɸcosɸ
xsinɸ . (sin^{2}ɸ + cos^{2}ɸ) = sinɸcosɸ
so xsinɸ = sinɸcosɸ
so x = cosɸ
now since xsinɸ = ycosɸ
solve by putting x = cosɸ, we get y = sinɸ
so x^{2} + y^{2} = 1 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeIf ABCDEF is a regular hexagon, then ∆ACE is
Correct
Incorrect

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeRam invested Rs 12585 in 3 parts which all are invested at 5% per annum. If he got same amount after 2, 4 and 5 yrs from 3 parts respectively, find the ratio in which the money was divided to be invested.
Correct
Explanation:
Ratio is
1/(100 + 2*5) : 1/(100 + 4*5) : 1/(100 + 5*5)
1/110 : 1/120 : 1/125Incorrect
Explanation:
Ratio is
1/(100 + 2*5) : 1/(100 + 4*5) : 1/(100 + 5*5)
1/110 : 1/120 : 1/125 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeFind the least number which when divided by 9, 12, 18 and 22 leaves remainders 6, 9, 15 and 19 respectively.
Correct
Explanation:
Since 96 = 3, 129 = 3, 1815 = 3, 2219 = 3
So answer will be the LCM of these numbers – 3
So 396 – 3Incorrect
Explanation:
Since 96 = 3, 129 = 3, 1815 = 3, 2219 = 3
So answer will be the LCM of these numbers – 3
So 396 – 3 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThere are 40 students in a hostel. The total expense of mess increased by Rs 50 per day when 10 more students joined the hostel. Due to this expenditure per head decreased by Re 1. What is the amount of original expenditure per day of mess?
Correct
Explanation:
Let E be original exp., then exp per head is E/40
Now E becomes (E+50) and students are 50 now, so exp per head is (E+50)/50
So, E/40 – (E+50)/50 = 1
Solve, E = 400Incorrect
Explanation:
Let E be original exp., then exp per head is E/40
Now E becomes (E+50) and students are 50 now, so exp per head is (E+50)/50
So, E/40 – (E+50)/50 = 1
Solve, E = 400
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