Direction[1-3]: What should come in place of the question mark (?) in the following number series?
1. 30 , 62 , 189 , ? , 3805 , 22836

A. 860 B. 760 C. 560 D. 555 E. 654

2. 1.5 , 2.5 , 6 , 19 , 77 , ?

A. 420 B. 99 C. 126 D. 386 E. 102

3. 140, 133, 119, 98, ?, 35

A. 70 B. 45 C. 30 D. 72 E. 61

4. On 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)

A. 28 B. 33 C. 46 D. 56 E. 37

5. Two trains for Mumbai leave Delhi at 6 am and 6:45 am and travel at 100 kmph and 136 kmph respectively. Approx. how many km from Delhi will the two trains be together?

A. 208 km B. 283 km C. 324 km D. 378 km E. None of these

Answers and Explanations
1. Answer - Option B
Explanation -
30 × 2 + 2 = 62
62 × 3 + 3 = 189
189 × 4 + 4 = 760
760 × 5 + 5 = 3805
3805 × 6 + 6 = 22836
30 , 62 , 189 , 760 , 3805 , 22836
Hence, option B is correct.
2. Answer - Option D
Explanation -

3. Answer - Option A
Explanation -
The Pattern is:-
140
140 - 7 = 133
133 - 14 = 119
119 - 21 = 98
98 - 28 = 70
70 - 35 = 35
Thus, the missing number is 70
So option (A) is the correct answer
4. Answer - Option E
Explanation -
Initially, the total age of the family is 6A (as average age of family is A and number of members in the family is 6)
Five years down the line the total age will increase by : 6*5
Then a child is born in the family and the total number of members now are 7, after a year total age will increase by 7*1
Given that, the average is still then A, so we can write the equation as:
6A + 6*5 + 7*1 = 7A ⇒ A = 37
5. Answer - Option B
Explanation -
In 45 min, 1st train will cover = [latex] 100 \times [\frac {45}{60}][/latex] = 75 km distance.
Relative speed of trains = 136 - 100 = 36 kmph.
2nd train will catch up the 1st train in = [latex]\frac {75}{36} = \frac {25}{12} [/latex]hours.
Distance from Delhi = [latex][\frac {25}{12}] \times 136 [/latex] = 283 km approx.

1. Six spherical cannon balls are tightly packed into a rectangular box in one layer. Each row has two cannon balls and each column has three. What part of the box is empty?

A. 11 B. [latex]\frac {10}{21}[/latex] C. [latex]\frac {15}{21}[/latex] D. 5 E. 20

2. K and L are two alloys of bronze and iron prepared by mixing the respective metals in the ratio of 5:3 and 5:11 respectively. If the alloys K and L are mixed to form a third alloy M with an equal ratio of bronze and iron then what is the ratio of alloys K and L in the new alloy M?

A. 3 : 5 B. 4 : 5 C. 3 : 2 D. 2 : 3 E. None of these

3. Siddhartha wants to buy a 11000 mah powerbank on a condition that he will pay 500 rs at the time of buying, 425 rs after 1 year and 289 rs after 2 years. If the compounded interest rate is 6 1/4 %, then what is the present value of powerbank?

A. 1150 B. 1050 C. 1156 D. 1256 E. 1080

Direction[4-5]: In the following number series, a wrong number is given. Find out that wrong number.
4. 7, 16, 24, 88, 113, 329

A. 88 B. 16 C. 113 D. 329 E. 444

5. 75, 85, 110, 135, 165, 200

A. 165 B. 85 C. 110 D. 200 E. 135

Answers and Explanations
1. Answer - Option B
Explanation -
Let the diameter of each ball be 2r.
Length of the box = 3 * 2r=6r
Breadth=2 * 2r = 4r
Height = 2r
Volume=6r * 4r * 2r = 48[latex]{r}^{3}[/latex]
Volume of 6 balls = [latex] 6 \times (\frac {4}{3}) \times (\frac {22}{7}) \times {r}^{3} = 176 \frac {{r}^{3}}{7}[/latex]
The area of empty space = [latex]48 {r}^{3} -176 \frac {{r}^{3}}{7}[/latex]
= [latex]160 \frac {{r}^{3}}{7}[/latex]
The required fraction = [latex]\frac {160 \frac {{r}^{3}}{7}}{48} = {r}^{3} = \frac {10}{21}[/latex]
2. Answer - Option C
3. Answer - Option C
Explanation -
Let the present value of powerbank = x rs.
After paying 500 rs. Rest cost = (x-500) rs.
∴ compounded amount of this cost for 1 year = (x - 500)[latex] (1 + \frac {\frac {25}{4}}{100})[/latex]
= [latex]\frac {17}{16}(x - 500)[/latex]
Now, rest cost after paying 425 rs. = [latex]\frac {17}{16}(x - 500)[/latex] - 425 rs
∴ compounded amount for next year =
[latex][\frac {17}{16}(x - 500) - 425 ] \times (1 + \frac {\frac {25}{4}}{100})[/latex]
= [latex]\frac {17}{16} [\frac {17}{16}(x - 500) - 425][/latex]rs.
Now, after paying 289 rs. All instalments are completed,
So
[latex]\frac {17}{16} [\frac {17}{16}(x - 500) - 425][/latex]rs. = 289
⇒17(x - 500) - 425 * 16 = 272 * 16
⇒17x – 8500 – 6800 = 4352
⇒17x = 19652
⇒x = 1156
⇒Present value of powerbank = 1156 rs.
4. Answer - Option B
Explanation -
Pattern is [latex] + {2}^{3}, + {3}^{2}, + {4}^{3}, + {5}^{2}, + {6}^{3}[/latex]
Hence wrong no. is 16
5. Answer - Option B
Explanation -
The series is:
75 + 3 * 5= 90
90 + 4 * 5= 110
110 + 5 * 5 = 135
135 + 6 * 5 = 165
165 + 7 * 5 = 200
Therefore 85 is wrong

Direction[1-5]: Read the line chart given below and answer the question that follow.

1. What is the difference between the total sale of Bartaman newspaper and the total sale of ABP newspaper in all the localities together?

A. 400 B. 360 C. 350 D. 300 E. None of the above

2. The sale of Bartaman newspaper in Bidhan Nagar is approximately what percent of the total sale of Bartaman newspaper in all the localities together?

A. 18% B. 20% C. 22% D. 15% E. 12%

3. What is the ratio of the sale of Bartaman newspaper in locality Dhiman Nagar to the sale of ABP newspaper in locality Chandan Nagar?

A. 13 : 12 B. 4 : 3 C. 5 : 4 D. 7 : 3 E. None of these

4. The sale of Bartaman and ABP newspaper in locality Ananda Nagar is what percent of the sale of the same newspaper in locality English bazar?

A. 72.33% B. 68.33% C. 50% D. 58% E. 60%

5. What is the average sale of Bartaman and ABP newspaper in locality Bidhan Nagar and Dhiman Nagar?

A. 4750 B. 4850 C. 5000 D. 2890 E. None of these

Answers and Explanations
1. Answer - Option D
Explanation -
Bartaman = 1500 + 2000 + 1800 + 3500 + 2500 = 11300
ABP = 2600 + 1000 + 1500 + 3000 + 3500 = 11600
Difference = 11600 - 11300 = 300.
2. Answer - Option A
Explanation -
Total Bartaman newspapers= 11300
Locality B’s percentage of Bartaman newspaper = [latex]\frac {2000}{11300} \times 100 [/latex]% = 17.6% ≈ 18%
3. Answer - Option D
Explanation -
Amount of Bartaman newspaper in Dhiman Nagar and Chandan Nagar are 3500 and 1500.
3500 : 1500
=35 : 15 = 7 : 3.
4. Answer - Option B
Explanation -
Both newspapers in English bazaar = 3500 + 2500 = 6000 ------100
Both newspaper in Ananda Nagar = 2600 + 1500 = 4100
The required answer = [latex]\frac {4100}{6000} \times 100 [/latex]% = 68.33%
5. Answer - Option A
Explanation -
Bidhan Nagar = (2000 + 1000) = 3000
Dhiman Nagar = (3000 + 3500) = 6500
The required answer, Average = [latex]\frac {(6500 + 3000 )}{2}[/latex]
= [latex]\frac {9500}{2}[/latex]
= 4750.