Would you like to submit the test? |

Sample Aptitude Test 6 for you to Practice. Evaluate your Aptitude Test 6 test answering skills by trying the online Aptitude Sample Test 6 and know your score.

You have scored 3 out of 10

You Scored:2

Total Questions:2

Attended Questions:2

Correct Answered:2

1. In the given series, can you find the Missing number ?

67, 58, 49, X , 31, 22

67, 58, 49, X , 31, 22

Explanation:

The given series is -- 67, 58, 49, X , 31, 22

Consecutive numbers decreases by 9

Following this pattern, the Missing Number X should decrease by 9 from the previous number.

When we subtract 9 to the number before X (i.e 49) we get 49--9=40

Consecutive numbers decreases by 9

Following this pattern, the Missing Number X should decrease by 9 from the previous number.

When we subtract 9 to the number before X (i.e 49) we get 49--9=40

2. What will be the volume of the cone if the radius and height are of 1 unit length.

3. What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 16 % per annum?

Explanation:

Given principal = 5000

No. of years = 2

Rate of interest = 16

**Amount = P x (1+r/100)^n,**

= 5000 x (1+16/100)^2

= 5000 x (1 + 4/25)^2

= 5000 x (29 / 25)^2

= 5000 x (841 / 625)

=6728

=>**Amount = Rs.6728
Compound Interest = Amount - Principal**

= 6728 - 5000

Thus,Compound Interest = Rs.1728

4. Can you find the approximate value for the below equation

50.99 - 55.01 + 39.99 / 40.01 * 31.99

50.99 - 55.01 + 39.99 / 40.01 * 31.99

Explanation:

Given Question is 50.99 - 55.01 + 39.99 / 40.01 * 31.99

when we apply BODMAS RULE and round it off, we get => 51 - 55 + ( 40/ 40 ) * 32 = 51 - 55 + 1* 32 = 28

when we apply BODMAS RULE and round it off, we get => 51 - 55 + ( 40/ 40 ) * 32 = 51 - 55 + 1* 32 = 28

5. In an arithmetic progression the first term is 6 and its common difference is 5. If the general term is a_{n} , find a_{6} - a_{4}.

Explanation:

Given a = 6 and common difference = 5.

**General Term a**_{n} = a +(n-1)d

a_{n} = 6 + 5(n-1) = 5n + 1

a_{6} = 30 + 1=31

a_{4} = 20 + 1 = 21

**a**_{6} - a_{4} = 31 - 21 = 10

a

a

a

6. Total marks obtained by a class is 720 and the average score of the class is 72. Find the total number of students in the class?

Explanation:

Total No. of students = (720 / 72) =

7. 44% of a number is 275, what is 64% of that number?

Explanation:

Let k be the number.

Given, 44 % of k = 275

=> 44k/100 = 275

=> 4k/100 = 25

=>k/25 = 25

=>**k = 625 **

64% of 625 = (64/100) x 625

= 64/4 x 25

= 16 x 25

=**400**

Given, 44 % of k = 275

=> 44k/100 = 275

=> 4k/100 = 25

=>k/25 = 25

=>

64% of 625 = (64/100) x 625

= 64/4 x 25

= 16 x 25

=

8. Can you find the answer for the below equation

43 / 166 - 159 * 162 + 155

43 / 166 - 159 * 162 + 155

Explanation:

Given Question is 43 / 166 - 159 * 162 + 155

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) ( 43 / 166 ) - 159 * 162 + 155 = 0.25-159*162+155 = -25602.74

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) ( 43 / 166 ) - 159 * 162 + 155 = 0.25-159*162+155 = -25602.74

9. Arun borrowed Rs. 1200 from his friend at the rate of 9 (1/2) % per annum of 3 years. Find the amount to be paid at the end of 3 years.

Explanation:

Given

Principal, p = 1200 Rs, Rate of interest, r = 9 (1/2) % = 19/2% Time, n= 3 years

**Simple Interest, S.I = ( p x n x r) / 100**

S.I = (1200 x 19 x 3) / (2 x 100) = 6 x 19 x 3 = 342

Amount to be paid =**Principal + S.I **= 1200 + 342 = **1542**

Amount to be paid at the end of 3 years =1542.

Principal, p = 1200 Rs, Rate of interest, r = 9 (1/2) % = 19/2% Time, n= 3 years

S.I = (1200 x 19 x 3) / (2 x 100) = 6 x 19 x 3 = 342

Amount to be paid =

Amount to be paid at the end of 3 years =1542.

10. A professor wrote the below series in classroom. He wants his students to find the next number. Can you solve it for the professor?

112, 91, 70, 49, _____

112, 91, 70, 49, _____

Explanation:

The given series is -- 112, 91, 70, 49

Consecutive numbers decreases by 21

Following this pattern, the Missing Number should decrease by 21 from the previous number.

When we subtract 21 to the last number (i.e 49) we get 49--21=28

Consecutive numbers decreases by 21

Following this pattern, the Missing Number should decrease by 21 from the previous number.

When we subtract 21 to the last number (i.e 49) we get 49--21=28

11. Can you find the answer for the below equation

58 / 158 - 154 + 155 * 150

58 / 158 - 154 + 155 * 150

Explanation:

Given Question is 58 / 158 - 154 + 155 * 150

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) ( 58 / 158 ) - 154 + 155 * 150 = 0.36-154+155*150 = 23096.37

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) ( 58 / 158 ) - 154 + 155 * 150 = 0.36-154+155*150 = 23096.37

12. Can you find the answer for the below equation

116 + 122 * 118 - 120 / 124

116 + 122 * 118 - 120 / 124

Explanation:

Given Question is 116 + 122 * 118 - 120 / 124

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) 116 + 122 * 118- ( 120/ 124 ) = 116 + 122 * 118- 0.96 = 14511.03

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) 116 + 122 * 118- ( 120/ 124 ) = 116 + 122 * 118- 0.96 = 14511.03

13. Evaluate the expression: 2 * 3 * 32 / 4 * 2 + 14 + 6

14. A fruit seller bought oranges at the rate of 5 for a rupee and sold them 4 for a rupee. The percent gain or loss is –

Explanation:

Let vendor buy 20 oranges (LCM of 5 and 4)
CP of 20 oranges = ?4
SP of 20 oranges = ?5
Gain % = (5 – 4)/4 x 100 = 25%

15. Can you find the the missing number in the below series

91, 84, 77, X , 63, 56

91, 84, 77, X , 63, 56

Explanation:

The given series is -- 91, 84, 77, X , 63, 56

Consecutive numbers decreases by 7

Following this pattern, the Missing Number X should decrease by 7 from the previous number.

When we subtract 7 to the number before X (i.e 77) we get 77--7=70

Consecutive numbers decreases by 7

Following this pattern, the Missing Number X should decrease by 7 from the previous number.

When we subtract 7 to the number before X (i.e 77) we get 77--7=70

16. In the first 35 overs of a cricket game, the run rate was only 6.4. What should be the run rate in the remaining 15 overs to reach the target of 410 runs. ?

Explanation:

"Required Run rate = ( 410 - ( 6.4 x 35 ) )/ (15)
= ( 186) /(15) = 12.4 "

17. 3x^{2} – 10x + 8 = 0,

3y^{2} + 14y + 16 = 0

3y

Explanation:

Given, 3x^{2} – 10x + 8 = 0

=> sum of roots = -10 => (-6 - 4 = -10)

Product of roots = 24 => (-6 * -4 = 24)

Thus the roots are -6, -4

=> 3x^{2} – 6x – 4x + 8 = 0

=> 3x(x - 2) - 4(x - 2) =0

=>(x - 2) (3x - 4) =0

=>x = 2, 4/3

**=> x = 2, 1.33**

Given, 3y^{2} + 14y + 16 = 0

=> sum of roots = 14 => (6 + 8 = 14)

Product of roots = 48 => (6 * 8 = 48)

Thus the roots are 6, 8

=> 3y^{2} + 6y + 8y + 16 = 0

=> 3y(y + 2) + 8(y + 2) =0

=>(y + 2) (3y + 8) = 0

=>y = -2, - 8/3

=>**y = - 2, -2.67**

Put on number line

**-2.67, -2, 1.33, 2**

Thus,**X > Y**

=> sum of roots = -10 => (-6 - 4 = -10)

Product of roots = 24 => (-6 * -4 = 24)

Thus the roots are -6, -4

=> 3x

=> 3x(x - 2) - 4(x - 2) =0

=>(x - 2) (3x - 4) =0

=>x = 2, 4/3

Given, 3y

=> sum of roots = 14 => (6 + 8 = 14)

Product of roots = 48 => (6 * 8 = 48)

Thus the roots are 6, 8

=> 3y

=> 3y(y + 2) + 8(y + 2) =0

=>(y + 2) (3y + 8) = 0

=>y = -2, - 8/3

=>

Put on number line

Thus,

18. There were 45 students in a hostel, if the numbers of students increased by 7, the expenses of the mess were increased by Rs.39 per day while the average expenditure per head diminished by Re.1. What is the original expenditure of the mess (in Rs)?

Explanation:

Let the original expenditure be Rs.X

Original average expenditure = X/45

New average expenditure = (X+39) / 52

(Since 7 students added, 45 + 7 = 52)

So (X/45) - ((X+39) / 52) = 1

=> (52X - 45X -1755) / 2340 = 1

=> 7X - 1755 = 2340

=> 7X =4095

=> X =4095 / 7

=>**X = 585**

So, original expenditure is**Rs. 585**

Original average expenditure = X/45

New average expenditure = (X+39) / 52

(Since 7 students added, 45 + 7 = 52)

So (X/45) - ((X+39) / 52) = 1

=> (52X - 45X -1755) / 2340 = 1

=> 7X - 1755 = 2340

=> 7X =4095

=> X =4095 / 7

=>

So, original expenditure is

19. What will be the unit's digit in the number N^2, if N = 1 + 2 + .. + 35

Explanation:

Given N = 1 + 2 + … + 35 = (35 x 36)/2 = 35 x 18
N is a number which ends with 0, hence N2 also has to end with 0.

20. Harvey walks in a Rectangle shaped park. Harvey observes the length and breadth to be 25m and 15m respectively. To plough the given Rectangle Park, a merchant charges Rs.18 per Sq.m. How much money have to be paid to the merchant?

Explanation:

Given Length = 25m & Breadth = 15m

Area of Rectangle = Length x Breadth => 25 x =15 = 375

Price = Area x Cost Per Sq.m = 375 x 18 = 6750

Area of Rectangle = Length x Breadth => 25 x =15 = 375

Price = Area x Cost Per Sq.m = 375 x 18 = 6750

21. Evaluate: 432 x 637 + 367 x 432 =?

Explanation:

432 x 637 + 363 x 432 =432 x ( 637 + 363) = 432 x 1000 = 432000

22. The largest 5 digit number exactly divisible by 91 is:

Explanation:

The largest 5 digit number is 99999

On dividing 99999 by 91, we get

=> Quotient =1098

=> Remainder = 81

So, 99999 – 81 =**99918**

**Thus the largest 5-digit number exactly divisible by 91 = ****99918**

On dividing 99999 by 91, we get

=> Quotient =1098

=> Remainder = 81

So, 99999 – 81 =

23. Can you find the approximate value for the below equation

102.99 / 113.99 * 100.01 - 107.01 + 96.01

102.99 / 113.99 * 100.01 - 107.01 + 96.01

Explanation:

Given Question is 102.99 / 113.99 * 100.01 - 107.01 + 96.01

when we apply BODMAS RULE and round it off, we get => ( 103 / 114 ) * 100 - 107 + 96 = 1*100-107+96 = 79

when we apply BODMAS RULE and round it off, we get => ( 103 / 114 ) * 100 - 107 + 96 = 1*100-107+96 = 79

24. Petrol & Kerosene are mixed in the ratio of 11:2. Total Volume of the mixture is 350 litres. Find the composition of Petrol in the overall mixture?

Explanation:

Given Petrol & Kerosene are in the ratio of 11:2
Given volume of mixture = 350 litres
Quantity of Petrol = 11/(11+2) x 350
= 296.15

25. Somu and Ramu can complete the work in 20 and 30 days respectively. In how many days 50 % of the work will get completed.

26. Can you find the the missing number in the below series

83, 76, 69, X , 55, 48

83, 76, 69, X , 55, 48

Explanation:

The given series is -- 83, 76, 69, X , 55, 48

Consecutive numbers decreases by 7

Following this pattern, the Missing Number X should decrease by 7 from the previous number.

When we subtract 7 to the number before X (i.e 69) we get 69--7=62

Consecutive numbers decreases by 7

Following this pattern, the Missing Number X should decrease by 7 from the previous number.

When we subtract 7 to the number before X (i.e 69) we get 69--7=62

27. Find the next number in the the below series

16, 30, 47, 67, _____

16, 30, 47, 67, _____

Explanation:

In the given series, 16, 30, 47, 67 ,

Consecutive numbers increase by 14 ,17 ,20.

Following the same pattern, the next number should increase by 23.

When we add 23 to the last number (i.e 67) we get 23+67=90

Consecutive numbers increase by 14 ,17 ,20.

Following the same pattern, the next number should increase by 23.

When we add 23 to the last number (i.e 67) we get 23+67=90

28. 9 men or 14 women can construct a wall in 23 days. How long will it take for 6 men and 6 women to do the same work?

Explanation:

Given that

**9 men** or **14 women** can construct a wall in **23 days. **

Let M denote men and W represent women, from the above statement we can say that

9M = 14 W or

M = (14/9) W

Now, the question is how many days 6 men and 6 women will take to complete the same work. which means, we have to evaluate

6M + 6W 6M + 6W

= 6(14 / 9)W + 6W

= 28W / 3 + 6W

=**46W / 3**

If 14 Women can complete a work in 23 days, then the time taken to complete the same work by 46W/3 will be

= (14 x 23 x 3) /46 days

= 966 / 46

=**21 days.**

Let M denote men and W represent women, from the above statement we can say that

9M = 14 W or

M = (14/9) W

Now, the question is how many days 6 men and 6 women will take to complete the same work. which means, we have to evaluate

6M + 6W 6M + 6W

= 6(14 / 9)W + 6W

= 28W / 3 + 6W

=

If 14 Women can complete a work in 23 days, then the time taken to complete the same work by 46W/3 will be

= (14 x 23 x 3) /46 days

= 966 / 46

=

29. Can you find the answer for the below equation

154 * 288 + 286 / 289 - 288

154 * 288 + 286 / 289 - 288

Explanation:

Given Question is 154 * 288 + 286 / 289 - 288

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) 154 * 288 + ( 286/ 289 ) - 288 = 154 * 288 + 0.98- 288 = 44064.99

when we apply BODMAS RULE to the above question, we get (Bracket Of Division) 154 * 288 + ( 286/ 289 ) - 288 = 154 * 288 + 0.98- 288 = 44064.99

30. In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of:

Explanation:

To obtain Rs. 10, investment = 96
For 650 Rs, investment = 96/10 x 650 = Rs. 6240