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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeTwo parallel chords of a circle of diameter 10 cm lying on the opposite sides of the centre are of lengths 6 cm and 10 cm. the distance between the chords is
Correct
Explanation:
OA = OC = 5 cm
AB = 6 cm, so AP = PB = 3 cm
CD = 8 cm, so CQ = QD = 4 cm
In ∆OCQ,
OQ = √(5^{2} – 4^{2}) = 3
In ∆OAP,
OP = √(5^{2} – 3^{2}) = 4
So PQ = OP+OQ = 4+3Incorrect
Explanation:
OA = OC = 5 cm
AB = 6 cm, so AP = PB = 3 cm
CD = 8 cm, so CQ = QD = 4 cm
In ∆OCQ,
OQ = √(5^{2} – 4^{2}) = 3
In ∆OAP,
OP = √(5^{2} – 3^{2}) = 4
So PQ = OP+OQ = 4+3 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeEvaluate 3 cos 70 cosec 20 + 5 cos 46 cosec 44
Correct
Explanation:
3 cos 70 cosec 20 + 5 cos 46 cosec 44
= 3 cos (9020) cosec 20 + 5 cos (9044) cosec 44
= 3 sin 20 cosec 20 + 5 sin 44 cosec 44
= 3 + 5Incorrect
Explanation:
3 cos 70 cosec 20 + 5 cos 46 cosec 44
= 3 cos (9020) cosec 20 + 5 cos (9044) cosec 44
= 3 sin 20 cosec 20 + 5 sin 44 cosec 44
= 3 + 5 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeThe ratio of 36^{3.5} : 6^{4} is
Correct
Explanation:
36^{3.5} : 6^{4}
(6^{2})^{7/2} : 6^{4}
6^{7} : 6^{4}
6^{3} : 1Incorrect
Explanation:
36^{3.5} : 6^{4}
(6^{2})^{7/2} : 6^{4}
6^{7} : 6^{4}
6^{3} : 1 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeIf angles of elevation of a tower from two consecutive kilometerstones along the road are 30 degree and 60 degree respectively, then the height of tower above the ground will be?
Correct
Explanation:
See the figure
tan 60 = h/BC so BC = h/√3
and tan 30 = h/(BC+1), so BC = √3 h – 1
so h/√3 = √3 h – 1
solve, h = √3/2Incorrect
Explanation:
See the figure
tan 60 = h/BC so BC = h/√3
and tan 30 = h/(BC+1), so BC = √3 h – 1
so h/√3 = √3 h – 1
solve, h = √3/2 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA price of an article is first decreased by 10% and then increased by 25%. If the final price of the article becomes Rs 360, what was the original price?
Correct
Explanation:
x * (90/100) * (125/100) = 360
solve, x = 320Incorrect
Explanation:
x * (90/100) * (125/100) = 360
solve, x = 320 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections (69): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.
The percentage passed in 1st division in 2013 was
Correct
Explanation:
Total students in 2013 = 170
Students passed in 1st division = 20
So required % = (20/170) * 100Incorrect
Explanation:
Total students in 2013 = 170
Students passed in 1st division = 20
So required % = (20/170) * 100 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections (69): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.
In which year the school had the best result in respect of percentage of pass candidates?
Correct
Explanation:
% of passed candidates:
Year 2013: 82 6/11%
Year 2014: (140/190)*100 = 73.7%
Year 2015: (150/200)8100 = 75%Incorrect
Explanation:
% of passed candidates:
Year 2013: 82 6/11%
Year 2014: (140/190)*100 = 73.7%
Year 2015: (150/200)8100 = 75% 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (69): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.
The number of students passed in third division in year 2014 was
Correct
Explanation:
Students passed in third division in 2014 = 190 – 140Incorrect
Explanation:
Students passed in third division in 2014 = 190 – 140 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections (69): Study the bar graph below and answer the following questions.
The graph shows the number of students and their evaluation for 3 years.
The pass percentage in 2013 was
Correct
Explanation:
Total passed students in 2013 is 140
So required% = (140/170)*100Incorrect
Explanation:
Total passed students in 2013 is 140
So required% = (140/170)*100 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe elimination of ɸ from x cosɸ – y sinɸ = 3 and x sinɸ + y sinɸ = 4 will give
Correct
Explanation:
Square both equations and add:
x^{2}cos^{2}ɸ + y^{2}sin^{2}ɸ – 2xy sinɸ cosɸ + x^{2}cos^{2}ɸ + y^{2}sin^{2}ɸ + 2xy sinɸ cosɸ = 9+16
x^{2}[cos^{2}ɸ + sin^{2}ɸ] + y^{2}[cos^{2}ɸ + sin^{2}ɸ] = 25
so x^{2} + y^{2} = 25Incorrect
Explanation:
Square both equations and add:
x^{2}cos^{2}ɸ + y^{2}sin^{2}ɸ – 2xy sinɸ cosɸ + x^{2}cos^{2}ɸ + y^{2}sin^{2}ɸ + 2xy sinɸ cosɸ = 9+16
x^{2}[cos^{2}ɸ + sin^{2}ɸ] + y^{2}[cos^{2}ɸ + sin^{2}ɸ] = 25
so x^{2} + y^{2} = 25
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