1. Can you find the the missing number in the below series
112, 85, 58, X , 4, -23
Answer - Option D
Explanation -
The given series is -- 112, 85, 58, X , 4, -23
Consequtive numbers decreases by 27
Following this pattern, the Missing Number X should decrease by 27 from the previous number.
When we subract 27 to the number before X (i.e 58)
we get 58 - 27 = 31
2. Find the Missing number in the the below series
112, 139, 166, X , 220, 247
A. 192
B. 191
C. 195
D. 193
Answer - Option D
Explanation -
The given series is -- 112, 139, 166, X , 220, 247
Consequtive numbers increases by 27
Following this pattern, the Missing Number X should increase by 27 from the previous number.
When we add 27 to the number before X (i.e 166) we get 166 + 27 = 193
3. A man borrows Rs. 12,500 at 20% compound interest. At the end of every year he pays Rs. 2000 as part of repayment. How much does he still owe after three such instalments?
A. 12, 000
B. 12, 864
C. 15, 600
D. 14, 320
Answer - Option D
Explanation -
Principal = Rs. 12,500
Rate = 20% compounded per annum.
WKT, Amount = [latex]P[1+ {(\frac {r} {100})}^{n} [/latex]
Amount after first year = [latex]12500 \times (\frac {120} {100}) = Rs. 15,000[/latex]
Principal for second year = 15000 - 2000 = Rs. 13,000
Amount after second year = [latex]13000 \times (\frac {120} {100}) = Rs. 15,600 [/latex]
Principal for third year = 15600 - 2000 = Rs. 13,600
Amount after third year = [latex]13600 \times (\frac {120} {100}) = Rs. 16,320[/latex]
Remaining amount = 16320 - 2000 = Rs. 14,320.
4. Subash purchased a refrigerator on the terms that he is required to pay Rs.1,500 as cash down payment followed by Rs.1,020 at the end of first year, Rs.1,003 at the end of second year and Rs.990 at the end of third year. Interest is charged at the rate of 10% per annum. Calculate the cost price.
A. 5200
B. 4000
C. 6500
D. 7800
Answer - Option B
Explanation -
Given
Cash Down Payment = Rs 1500
let X becomes 1020 at the end of 1st year
[latex]1020 = x ( 1 + (\frac {10}{100}) )[/latex]
x = [latex]1020 \times \frac {10}{11}[/latex]
[latex] \Rightarrow [/latex] 927.27
It becomes 1003 at the end of 2nd year
y = [latex] \frac {1003 \times 20 \times 20}{ 22 \times 22}[/latex]
[latex]\Rightarrow[/latex] 828.92
It becomes 990 at the end of 3rd year
z [latex]= 990 \times 10 \times 10 \times \frac {10}{11 \times 11 \times 11 }[/latex]
[latex]\Rightarrow[/latex] 743.80
Cost Price [latex]\Rightarrow[/latex] 1500 + 927.27 + 828.92 + 743.80
[latex]\Rightarrow[/latex] 3999.99 or 4000.
5. A sum of Rs. 700 is lent at the rate of 6% per annum. Find the interest at the end of 4 years.
A. 168
B. 158
C. 188
D. 198
Answer - Option A
Explanation -
Given principal, p = 700 Rs,
rate of interest r= 6%,
time, t = 4 years
Simple interest, SI = [latex]\frac {(p x r x t)} {100} [/latex]
[latex]\Rightarrow SI = \frac {(700 x 6 x 4)}{100}[/latex]
[latex]\Rightarrow[/latex] SI = 168
[latex]\Rightarrow[/latex] At the end of 4 years, interest earned is Rs. 168
6. A sum of Rs. 1092 is lent at the rate of 10% per annum. Find the interest at the end of 7 years.
A. 618.3
B. 943.2
C. 892.6
D. 764.4
Answer - Option D
Explanation -
Given principal, p = 1092 Rs,
rate of interest r= 10%,
time, n = 7 years
Simple interest, SI = [latex]\frac {(p x n x r)}{100} [/latex]
=> SI = [latex] \frac {(1092 x 7 x 10)}{100} [/latex]
=> SI = 764.4
At the end of 7 years, interest earned is Rs. 764.4
7. In a sale, perfume are available at a discount of 25% on the selling price. If a perfume costs Rs. 5895 in the sale, what is the selling price of the perfume?
A. Rs 6020
B. Rs 7860
C. Rs. 7680
D. 7000
Answer - Option B
Explanation -
Given : CP - Rs. 5895; Discount - 25%
Let the SP of perfume be 'X'.
CP = SP - 25% of SP
5895 = X - 25% of X
5895 = 75% of X
X = [latex] \frac {(5895 \times 100)}{75} [/latex]
X = Rs. 7860
Therefore, SP of perfume is Rs. 7860.
8. Dev got a discount of 30% on the marked price of an article which he sold for Rs. 8750 with 25% profit on the price he bought. Find the marked price?
A. 11,000
B. 12,000
C. 10,000
D. 9870
Answer - Option C
Explanation -
SP = Rs.8750; Profit =25%; Discount = 30%
WKT, CP = [latex]
SP[\frac {100}{(100 + Profit)}][/latex]
CP = [latex]8750 [\frac {100}{(100 + 25)]}[/latex]
[latex]= 8750[\frac {100}{125}][/latex]
= Rs.7000
As per the question discount is 30%, so
70% of MP = CP
[latex](\frac {70}{100})[/latex] x MP = CP
MP =[latex] 7000 x (\frac {100}{70})[/latex]
MP = Rs. 10,000.
9. Find the volume of the cylinder of radius 10 cm and height 35 cm.
A. 11000
B. 11250
C. 11500
D. 11550
Answer - Option A
Explanation -
Given,radius, r = 10 cm and
height, h = 35 cm.
Volume of the cylinder = [latex]pi \times r^2 \times h [/latex]
= [latex](\frac {22}{7}) \times 10^2 \times 35[/latex]
= [latex]22 \times 100 \times 5[/latex]
= 11,000
Therefore, Volume of Cylinder = 11,000 [latex]{cm}^{3}[/latex]
10. Given a trapezium of two parallel sides of length 43 and 57 cm, if the distance between the parallel side is 10 cm, find the area of the trapezium
A. 500
B. 520
C. 480
D. 400
Answer - Option A
11. Vijay requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum marks he could have got?
A. 500
B. 400
C. 450
D. 600
Answer - Option A
Explanation -
If Vijay had 15 marks, more he could have secured 40 % marks.
Now, 15 marks more than 185 = 185 + 15 = 200
Let the maximum marks be k, then 40 % of m =[latex] 200 \frac {40}{100} x m = 200 m = 500[/latex], thus the maximum mark is 500.
12. Raja earns 22 % of his investment. If he earns Rs. 187, then how much did he invest ?
A. 775
B. 750
C. 840
D. 850
Answer - Option D
Explanation -
Let the investment be Rs x
[latex]\Rightarrow[/latex] 22% of x = 187
[latex]\Rightarrow ( \frac {22}{100}) \times x = 187[/latex]
[latex]\Rightarrow x = 187 \times \frac {100}{22}[/latex]
[latex]\Rightarrow \frac {18700}{22} [/latex]
[latex]\Rightarrow[/latex] Rs 850.
[latex]\Rightarrow[/latex]Therefore , Radha invests Rs 850
13. Ratio between two numbers is 3:2 and their difference is 225, then the smaller number is:
A. 90
B. 675
C. 135
D. 450
Answer - Option D
Explanation -
Given, the ratio of two numbers = 3 : 2
Let the two numbers be 3x and 2x
Given, their difference = 225 [latex]\Rightarrow[/latex] 3x - 2x = 225 [latex]\Rightarrow[/latex] x = 225
Smaller number = 2x = 2 [latex]\times[/latex] 225 = 450
14. Two numbers are in the ratio 7:9. If 12 is subtracted from each of them, the ratio becomes 3:5. The product of the numbers is
A. 432
B. 567
C. 1575
D. 1263
Answer - Option B
Explanation -
Let the investment be Rs x
[latex]\Rightarrow [/latex] 22% of x = 187
[latex]\Rightarrow ( \frac {22}{100}) \times x = 187[/latex]
Given, ratio of two numbers = 7: 9
Let two numbers be 7x and 9x
If 12 is subtracted from each number,
[latex]\Rightarrow \frac {(7x -12)}{(9x - 12)} = \frac {3}{5} [/latex]
[latex]\Rightarrow [/latex] 5(7x - 12) = 3(9x - 12)
[latex]\Rightarrow [/latex] 35x - 60 = 27x - 36
[latex]\Rightarrow [/latex] 8x = 24
[latex]\Rightarrow [/latex] x = 3
Product of the numbers = 7x × 9x
= 7(3) × 9(3)
= 21 × 27
= 567
15. 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)?
A. 562
B. 585
C. 650
D. 548
Answer - Option B
Explanation -
Let the original expenditure be Rs.x
Original average expenditure = X/45
New average expenditure = [latex]\frac {(x+39)}{52}[/latex]
So [latex](\frac {x}{45}) - (\frac {(x+39)}{ 52}) = 1 \Rightarrow \frac {(52x - 45x - 45 \times 39)} { (45 * 52)} = 1 \Rightarrow \frac {(7x - 1755)} {2340} = 1 \Rightarrow 7x = 2340 + 1755 \Rightarrow[/latex] 7x = 4095
so x = 585
So, original expenditure is Rs 585
16. The average of 14 numbers is zero. Of them, at the most how many may be greater than zero ?
Answer - Option B
Explanation -
Average of 14 numbers is = 0
Therefore sum of all 14 numbers = 0
It is quite possible that 13 of these numbers may be positive and if their sum is "a" then 14th number is "(-a)"
17. In the first 10 overs of a cricket game, the run rate was only 5.6. What should be the run rate in the remaining 40 overs to reach the target of 254 runs?
A. 5.445
B. 5.94
C. 4.95
D. 8.525
Answer - Option C
Explanation -
Given
Target = 254 runs
First 10 overs, run rate = 5.6
Required Run rate
= [latex] \frac {( 254 - ( 5.6 x 10 )} {(40)}[/latex]
= [latex]\frac {(198 )}{(40)}[/latex]
= 4.95
18. Two taps A and B can fill the tank in 12 and 15 hours respectively. If both the taps are opened together, find the time taken to fill the tank (in minutes).
A. 410
B. 450
C. 400
D. 460
Answer - Option C
Explanation -
Tap A fill the tank in 12 hours
Tap B fill the tank in 15 hours
To find the Time taken by both taps opened together to fill the tank:
[latex]\frac {1}{(A + B)} = (\frac {1}{A}) + (\frac {1}{B}) [/latex]
[latex]\Rightarrow \frac {1}{(A + B)} = (\frac {1}{12}) + (\frac {1}{15}) [/latex]
[latex] \Rightarrow \frac {1}{(A + B)} = (\frac {27}{180}) [/latex]
Taking reciprocal on both sides
[latex]\Rightarrow[/latex] A + B = [latex] \frac {180} {27}[/latex]
[latex]\Rightarrow[/latex] A + B = [latex]\frac {20}{3} hour \rightarrow [/latex] To convert hour into minute
[latex]\Rightarrow[/latex] A + B = [latex](\frac {20}{3}) \times 60[/latex] min
[latex]\Rightarrow[/latex]A + B = 400 mins
19. A, B and C can do a work in 5 days, 10 days and 15 days respectively. They started together to do the work but after 2 days A and B left. C did the remaining work (in days)
A. 2 days
B. 4 days
C. 5 days
D. 6 days
Answer - Option B
Explanation -
A's one day work = [latex]\frac {1}{5}[/latex], B's one day work [latex]\frac {1}{10}[/latex] and C's one day work [latex]\frac {1}{15}[/latex] If A, B and C work together for a day then they will finish [latex](\frac {1}{5} + \frac {1}{10} + \frac {1}{15}) = \frac {11}{30}[/latex] work. Therefore working together for two days they will finish [latex]2 \times \frac {11}{30} = \frac {11}{15}[/latex] work. C alone does remain [latex]1 - \times \frac {11}{15} = \frac {4}{15}[/latex] work. But C finishes [latex]\frac {1}{15}[/latex] work in one day. Therefore C will finish [latex]\frac {4}{15}[/latex] work in 4 days.
20. Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while C and A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job?
A. 4 days
B. 3 days
C. 6 days
D. 5 days
Answer - Option D
Direction(21-25): Refer to the table and answer the given questions.
Data related to number of employees in 5 different companies in December, 2012
Company |
Total number of Employees |
Out of total number of employees |
Percentage of Engineering Graduates |
Percentage of Science Graduates |
Percentage of Arts Graduates |
A |
1000 |
32% |
- |
- |
B |
600 |
- |
42% |
30% |
C |
- |
30% |
30% |
- |
D |
- |
- |
40% |
20% |
E |
- |
35% |
50% |
- |
I. Employees of the given companies can be categorised only in three types:
Engineering Graduates, Science Graduates and Arts Graduates
II. Few values are missing in the table (indicated by ___). A candidate is expected to calculate the missing value, if it is required to answer the given questions, on the basis of the given data and information.
21. What is the difference between the number of Arts graduate employees and Engineering Graduates employees in Company B?
Answer - Option C
Explanation -
Number of Arts graduate employees = 30% of 600 = 180
Number of Engineering Graduates employees = 28% of 600 = 168
Difference = 180 – 168 = 12
22. The Average number of Arts graduate employees and Science graduate employees in Company E was 338. what was the total number of employees in Company E?
A. 920
B. 960
C. 1120
D. 1040
Answer - Option D
Explanation -
Average no of Arts graduate employees and Science graduate employees in Company E = 338
Total no of Arts graduate employees and Science graduate employees in Company E = 676
Total number of employees in Company E = X
X = [latex] 676 \times* (\frac {100}{65}) = 1040[/latex]
23. If the respective ratio between the number of Science graduate employees and Arts graduate employees in Company A was 10:7, what was the number of Arts graduate employees in A?
A. 294
B. 266
C. 280
D. 308
Answer - Option C
Explanation -
Number of Science Graduate employees = 32% of 1000 = 320
Number of Science Graduate employees and Arts graduate employees = 1000 -320 = 680
Number of Arts graduate employees in A = 680 * [latex](\frac {7}{17})[/latex] = 280
24. Total number of employees in Company B increased by 20% from December 2012 to 2013, If 20% of the total number of employees in Company B in December, 2013 was Engineering graduates, what was the number of Engineering graduate employees in Company B in December 2013?
A. 194
B. 166
C. 144
D. 152
Answer - Option C
Explanation -
Total number of employees in Company B in 2013 = 720
Engineering graduate employees in Company B in December 2013 = 20% of 720 = 144
25. Total number of employees in Company C was 3 times the total number of employees in Company D. If the difference between number of Arts graduate employees in Company D and that of Science graduate employees in the same Company was 120, what was the total number of employees in Company C?
A. 1200
B. 1400
C. 1800
D. 1600
Answer - Option D
Explanation -
40% – 20% = 120
20% = 120; Total number of employees in Company D = 600
Total number of employees in Company C = 1800