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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

If √[(x-a)/(x-b)] + a/x = √[(x-b)/(x-a)] + b/x, and given that b ≠ a, then the value of x is

Correct

Explanation:
√[(x-a)/(x-b)] + a/x = √[(x-b)/(x-a)] + b/x
√[(x-a)/(x-b)] – √[(x-b)/(x-a)] = b/x – a/x
√[(x-a)(x-a)/(x-a)(x-b)] – √[(x-b)(x-b)/(x-a)(x-b)] = (b-a)/x
(x-a)/√[(x-a)(x-b)] – (x-b)/√[(x-a)(x-b)] = (b-a)/x
(x-a – x + b)/√[(x-a)(x-b)] = (b-a)/x
So x = √[(x-a)(x-b)]
Square both sides
x^2 = (x-a)(x-b)
solve, ax + bx = ab
so x = ab/(a+b)

Incorrect

Explanation:
√[(x-a)/(x-b)] + a/x = √[(x-b)/(x-a)] + b/x
√[(x-a)/(x-b)] – √[(x-b)/(x-a)] = b/x – a/x
√[(x-a)(x-a)/(x-a)(x-b)] – √[(x-b)(x-b)/(x-a)(x-b)] = (b-a)/x
(x-a)/√[(x-a)(x-b)] – (x-b)/√[(x-a)(x-b)] = (b-a)/x
(x-a – x + b)/√[(x-a)(x-b)] = (b-a)/x
So x = √[(x-a)(x-b)]
Square both sides
x^2 = (x-a)(x-b)
solve, ax + bx = ab
so x = ab/(a+b)

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

If 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) = 2

Incorrect

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 points

Category: Quantitative Aptitude

A 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 * 18000

Incorrect

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 points

Category: Quantitative Aptitude

Two 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(20-10) – 2*20*10] / [(100+20) + (100-10)] = + 2 6/7%

Incorrect

Explanation:
Use formula
[100(20-10) – 2*20*10] / [(100+20) + (100-10)] = + 2 6/7%

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

ABCD 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 = (10-6)/2 = 2
So CD = √(DF^2 +FC^2) = √(8^2 + 2^2) = 2√17

Incorrect

Explanation:

Area of trap = ½ * (AD+BC) * DF
64 = 1/2*(6+10) * DF
So DF = 8
FC = (10-6)/2 = 2
So CD = √(DF^2 +FC^2) = √(8^2 + 2^2) = 2√17

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

If xsin^{3}ɸ + ycos^{3}ɸ = sinɸcosɸ ≠ 0 and xsinɸ – ycosɸ = 0, then value of (x^{2} + y^{2}) =

xsinɸ . 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

xsinɸ . 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 points

Category: Quantitative Aptitude

If ABCDEF is a regular hexagon, then ∆ACE is

Correct

Incorrect

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Ram 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 : 1/125

Incorrect

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 points

Category: Quantitative Aptitude

Find the least number which when divided by 9, 12, 18 and 22 leaves remainders 6, 9, 15 and 19 respectively.

Correct

Explanation:
Since 9-6 = 3, 12-9 = 3, 18-15 = 3, 22-19 = 3
So answer will be the LCM of these numbers – 3
So 396 – 3

Incorrect

Explanation:
Since 9-6 = 3, 12-9 = 3, 18-15 = 3, 22-19 = 3
So answer will be the LCM of these numbers – 3
So 396 – 3

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

There 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 = 400

Incorrect

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