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Question 1 of 10
1. Question
1 pointsCategory: Quantitative Aptitude[(5√6 + √6) x (16√6 + 9√6)] – 96 = ?
Correct
Answer : 2. 804
Explanation :
[(5√6 + √6) x (16√6 + 9√6)] – 96
[√6(5 + 1) x √6(16 + 9)] – 96
900 – 96 = 804Incorrect
Answer : 2. 804
Explanation :
[(5√6 + √6) x (16√6 + 9√6)] – 96
[√6(5 + 1) x √6(16 + 9)] – 96
900 – 96 = 804 -
Question 2 of 10
2. Question
1 pointsCategory: Quantitative Aptitude√(49 x 24 x 2 -(11)³ + 68) = ?
Correct
Answer : 2. 33
Explanation :
√(2352 – 1331 + 68) = √1089 =33Incorrect
Answer : 2. 33
Explanation :
√(2352 – 1331 + 68) = √1089 =33 -
Question 3 of 10
3. Question
1 pointsCategory: Quantitative Aptitude14, 27, 43, 65, ?, 157
Correct
Answer : 1. 99
Explanation :
14*2 – 1 = 27
27*2 – 11 = 43
43*2 – 21 = 65Incorrect
Answer : 1. 99
Explanation :
14*2 – 1 = 27
27*2 – 11 = 43
43*2 – 21 = 65 -
Question 4 of 10
4. Question
1 pointsCategory: Quantitative Aptitude4, ?, 120, 595, 2376 7125
Correct
Answer : 3. 21
Explanation :
4*7 -7 = 21
21*6 -6 = 120
120*5 -5 = 595.Incorrect
Answer : 3. 21
Explanation :
4*7 -7 = 21
21*6 -6 = 120
120*5 -5 = 595. -
Question 5 of 10
5. Question
1 pointsCategory: Quantitative Aptitudex² – 39x + 324 = 0
y² – 47y + 420 = 0Correct
Answer : 5. X = Y or relation cannot be established
Explanation:
x² – 39x + 324 = 0
x = 27, 12
y² – 47y + 420 = 0
y = 35, 12Incorrect
Answer : 5. X = Y or relation cannot be established
Explanation:
x² – 39x + 324 = 0
x = 27, 12
y² – 47y + 420 = 0
y = 35, 12 -
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeAjay’s present age is 1.5 times of Hardik’s present age. If after 4 years, Ajay’s age will be twice of Hardik’s age 4 years ago. What is the sum of the present ages of Ajay and Hardik?
Correct
Answer : 3. 60 years
Explanation :
A = 1.5 * H
A + 4 = 2 * (H – 4)
A + 4 = 2H – 8
A + 12 = 2H
1.5H + 12 = 2H
H = 24
A = 24 * 1.5 = 36
Sum = 36 + 24 = 60Incorrect
Answer : 3. 60 years
Explanation :
A = 1.5 * H
A + 4 = 2 * (H – 4)
A + 4 = 2H – 8
A + 12 = 2H
1.5H + 12 = 2H
H = 24
A = 24 * 1.5 = 36
Sum = 36 + 24 = 60 -
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe cost of servicing of a Etios car at one of the Toyota Service Centers is Rs. 400. Manager of service center told that for the second service within a year a customer can avail a 10% discount and further for third and forth servicing the customer can avail 10% discount of the previous amount paid, within a year Further if a customer gets more than 4 services within a year he has to pay just 60% of the servicing charges on these services. Vijay availed 5 services from the same servicing station, What is the total percentage discount fetched by Vijay?
Correct
Answer : 1. 19.42%
Explanation :
Let Amount paid in 1st service = 100
Amount paid in 2nd service = 90
Amount paid in 3rd service = 81
Amount paid in 4th service = 72.9
Amount paid in 5th service = 60
Total amount paid = 403.9
Discount = 500 – 403.9 = 96.1
Discount % = (96.1 / 500) * 100 = 19.42%Incorrect
Answer : 1. 19.42%
Explanation :
Let Amount paid in 1st service = 100
Amount paid in 2nd service = 90
Amount paid in 3rd service = 81
Amount paid in 4th service = 72.9
Amount paid in 5th service = 60
Total amount paid = 403.9
Discount = 500 – 403.9 = 96.1
Discount % = (96.1 / 500) * 100 = 19.42% -
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeA boat’s speed in still water is 5 kmph. While river is flowing with a speed of 2 kmph and time taken to cover a certain distance upstream is 4 hours more than time taken to cover the same distance downstream. Find the distance?
Correct
Answer : 2. 21 km
Let the distance be D km.
Speed download = (5+2) = 7 km/h
and speed upstream = (5-2) = 3 km/h
According to the question,
D/3 – D/7 = 4
⇒ 7D – 3D = 21 x 4
D = (21 x 4)/4 = 21 kmIncorrect
Answer : 2. 21 km
Let the distance be D km.
Speed download = (5+2) = 7 km/h
and speed upstream = (5-2) = 3 km/h
According to the question,
D/3 – D/7 = 4
⇒ 7D – 3D = 21 x 4
D = (21 x 4)/4 = 21 km -
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeOut of the three semester examinations, each with a total of 500 marks, a student secured average marks of 45% and 55% in the first and second semester examinations. To have an overall average of 60%, how many marks does the student need to secure in the third semester examination?
Correct
Answer : 2. 400
Explanation :
Total marks for three semester examinations = 3x 500 = 1500
Total required marks in three semester examinations = 60% of 1500
=( 3 x 500 x 60 ) / 100 = 900
Marks secured in first semester examination = 45 % of 500
=( 500 x 45 ) / 100 =225
Marks secured in third semester examination = 55 % of 500
=( 500 x 55 ) / 100 =275
Thus, the required marks in third semester examination
=900 – ( 225 + 275 ) =900 – 500 = 400Incorrect
Answer : 2. 400
Explanation :
Total marks for three semester examinations = 3x 500 = 1500
Total required marks in three semester examinations = 60% of 1500
=( 3 x 500 x 60 ) / 100 = 900
Marks secured in first semester examination = 45 % of 500
=( 500 x 45 ) / 100 =225
Marks secured in third semester examination = 55 % of 500
=( 500 x 55 ) / 100 =275
Thus, the required marks in third semester examination
=900 – ( 225 + 275 ) =900 – 500 = 400 -
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeAn aspirant attempted 12 questions in a competitive exam and secured full marks in all of them. If he obtained 60% in the competitive exam and each questions carry equal marks, then what was the total number of questions in the competitive exam?
Correct
Answer : 4. 20
Explanation :
Let the number of question be Y.
60% of Y = 12
60Y / 100 =12
Y = 20Incorrect
Answer : 4. 20
Explanation :
Let the number of question be Y.
60% of Y = 12
60Y / 100 =12
Y = 20
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