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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA boy has a piggybank in which there are 10, 25 and 50 paise coins. The coins are in ratio 4 : 3 : 5 respectively. If the total amount in piggybank is Rs 109.50 then what is the total number of coins in the piggybank?
Correct
Explanation:
109.50 rs = 10950 paise
Now 4x, 3x, and 5x
So 4x*10 + 3x*25 + 5x*50 = 10950
Solve, x = 30
So total coins = 4x+3x+5xIncorrect
Explanation:
109.50 rs = 10950 paise
Now 4x, 3x, and 5x
So 4x*10 + 3x*25 + 5x*50 = 10950
Solve, x = 30
So total coins = 4x+3x+5x 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeIf the approximate value of 1^{3} + 2^{3} …. + 6^{3} = 441, then approximate value of (0.11)^{3} + (0.22)^{3} …. + (0.66)^{3} =
Correct
Explanation:
(0.11)^{3} + (0.22)^{3} …. + (0.66)^{3}
= 0.11^{3} (1^{3} + 2^{3} …. + 6^{3})
= 0.11*0.11*0.11*441 = 0.6Incorrect
Explanation:
(0.11)^{3} + (0.22)^{3} …. + (0.66)^{3}
= 0.11^{3} (1^{3} + 2^{3} …. + 6^{3})
= 0.11*0.11*0.11*441 = 0.6 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA takes 8 hr in riding at certain place and return back. While it takes 9 hrs in walking at certain place and riding back. Find the time A will take to walk both sides
Correct
Explanation:
R + R = 8, so R = 4
W+ R = 9, so W =5
So to walk both direction, he will take W+W = 5+5 = 10 hrs.Incorrect
Explanation:
R + R = 8, so R = 4
W+ R = 9, so W =5
So to walk both direction, he will take W+W = 5+5 = 10 hrs. 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeAB is a diameter of the circumcircle of ∆ACB. O is the foot of perpendicular drawn from the point C on AB. If AC = 4cm and BC = 3cm, then the length of OB is
Correct
Explanation:
Use formula
AC = 4cm and BC = 3cm
Angle ACB is 90
So AB^2 = 4^2 + 3^2
So AB = 5cm
Now in ∆OCB
CB^2 = BO^2 + OC^2
And in ∆OCA
CA^2 = AO^2 + OC^2
So
CB^2 – BO^2 = CA^2 – AO^2
Let BO = x, so AO = AB – BO = 5x
Now from CB^2 – BO^2 = CA^2 – AO^2
3^2 – x^2 = 4^2 – (5x)^2
Solve, x = 1.8Incorrect
Explanation:
Use formula
AC = 4cm and BC = 3cm
Angle ACB is 90
So AB^2 = 4^2 + 3^2
So AB = 5cm
Now in ∆OCB
CB^2 = BO^2 + OC^2
And in ∆OCA
CA^2 = AO^2 + OC^2
So
CB^2 – BO^2 = CA^2 – AO^2
Let BO = x, so AO = AB – BO = 5x
Now from CB^2 – BO^2 = CA^2 – AO^2
3^2 – x^2 = 4^2 – (5x)^2
Solve, x = 1.8 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeFind the value of (sinɸ – cosɸ + 1)/(sinɸ + cosɸ – 1) when ɸ ≠ ᴨ/2
Correct
Explanation:
(sinɸ – (cosɸ – 1))/(sinɸ + (cosɸ – 1))
So [(sinɸ – (cosɸ – 1))/(sinɸ + (cosɸ – 1))] * [(sinɸ – (cosɸ – 1))/ (sinɸ – (cosɸ – 1))]
=[Sin^{2}ɸ + cos^{2}ɸ – 2cosɸ + 1 – 2sinɸ(cosɸ1)]/[sin^{2}ɸ – cos^{2}ɸ + 2cosɸ 1]
= [2 – 2cosɸ + 2sinɸ – 2 sinɸcosɸ]/ [2cos^{2}ɸ + 2cosɸ]
= [2(1 – cosɸ) + 2sinɸ(1 – cosɸ)]/ [2cosɸ(1 – cosɸ)]
= 2(1 + sinɸ) (1 – cosɸ)/ [2cosɸ(1 – cosɸ)] = (1 + sinɸ)/cosɸIncorrect
Explanation:
(sinɸ – (cosɸ – 1))/(sinɸ + (cosɸ – 1))
So [(sinɸ – (cosɸ – 1))/(sinɸ + (cosɸ – 1))] * [(sinɸ – (cosɸ – 1))/ (sinɸ – (cosɸ – 1))]
=[Sin^{2}ɸ + cos^{2}ɸ – 2cosɸ + 1 – 2sinɸ(cosɸ1)]/[sin^{2}ɸ – cos^{2}ɸ + 2cosɸ 1]
= [2 – 2cosɸ + 2sinɸ – 2 sinɸcosɸ]/ [2cos^{2}ɸ + 2cosɸ]
= [2(1 – cosɸ) + 2sinɸ(1 – cosɸ)]/ [2cosɸ(1 – cosɸ)]
= 2(1 + sinɸ) (1 – cosɸ)/ [2cosɸ(1 – cosɸ)] = (1 + sinɸ)/cosɸ 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe angles of elevation of a top of tower standing on a horizontal plane from two points on a line passing through the foot of the tower at a distance 24ft and 54 ft respectively are complementary angles. Find the height of the tower (in ft).
Correct
Explanation:
In ∆ABD
tanɸ = AB/BD = AB/54
In ∆ABC
Tan(90ɸ) = AB/BC = AB/24
So cotɸ = AB/24
So tanɸ. cotɸ = AB/24 * AB/54
So AB^2 = 24*54
So AB = 36Incorrect
Explanation:
In ∆ABD
tanɸ = AB/BD = AB/54
In ∆ABC
Tan(90ɸ) = AB/BC = AB/24
So cotɸ = AB/24
So tanɸ. cotɸ = AB/24 * AB/54
So AB^2 = 24*54
So AB = 36 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeA metal pipe of negligible thickness has radius 9 cm and length 42 cm. The outer curved surface area of the pipe in sq. cm is
Correct
Explanation:
Curved surface area = 2ᴨrh = 2*(22/7)*9*42Incorrect
Explanation:
Curved surface area = 2ᴨrh = 2*(22/7)*9*42 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe radius of the circumcircle of the triangle made by xaxis, yaxis and 2x + 5y = 20 is
Correct
Explanation:
Since angle AOB = 90, so AB is diameter
AB^2 = AO^2 + OB^2 = 4^2 + 10^2 = 16+100 = 116
So radius = AB/2 = √29Incorrect
Explanation:
Since angle AOB = 90, so AB is diameter
AB^2 = AO^2 + OB^2 = 4^2 + 10^2 = 16+100 = 116
So radius = AB/2 = √29 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeAn article is bought at 10% discount on the marked price and sold at 20% profit on the marked price. Find the overall profit % in this transaction.
Correct
Explanation:
See that profit is on marked price here, not on cost price.
So let MP = 100 Rs
So CP = (10010)/100 * 100 = Rs 90
And SP = (100+20)/100 * 100 = Rs 120
So profit% = 30/90 * 100
OR shortcut (for this case only when profit on MP):
[(100+20) – (10010)]/(10010) * 100 = 30/90 * 100Incorrect
Explanation:
See that profit is on marked price here, not on cost price.
So let MP = 100 Rs
So CP = (10010)/100 * 100 = Rs 90
And SP = (100+20)/100 * 100 = Rs 120
So profit% = 30/90 * 100
OR shortcut (for this case only when profit on MP):
[(100+20) – (10010)]/(10010) * 100 = 30/90 * 100 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe total cost of painting the walls of a room is Rs 224. Find the cost of painting the walls of another room whose length, breadth and height each are two and a half times the dimensions of the previous room.
Correct
Explanation:
Area of first room = 2(l+b)*h
After all dimensions doubled, new area = 2(2.5l+2.5b)*2.5h = 6.25[2(l+b)*h ] = 6.25 times previous area, so cost of painting = 6.25*475Incorrect
Explanation:
Area of first room = 2(l+b)*h
After all dimensions doubled, new area = 2(2.5l+2.5b)*2.5h = 6.25[2(l+b)*h ] = 6.25 times previous area, so cost of painting = 6.25*475
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