1. In a class of 55 students there are 34 girls. The average weight of these girls is 51 Kg and average weight of the full class is 55.2 kgs. What is the average weight of the boys of the class?

A. 62 B. 59.4 C. 56.8 D. 60

Answer - Option A
Explanation -
Average weight of the boys = [latex]\frac{sum of weights of full class - sum of weights of girls}{number of boys}[/latex]
= [latex]\frac{55.2 \times 55 - 34 \times 51}{55 - 34}[/latex]
= [latex]\frac{3036 - 1734}{21}[/latex]
= 62 kg
2. The average revenues of 11 consecutive years of a company is Rs 69 lakhs. If the average of first 6 years is Rs 64 lakhs and that of last 6 years is Rs 76 lakhs, what is the revenue for the 6th year?

A. Rs 83 lakhs B. Rs 79 lakhs C. Rs 77 lakhs D. Rs 81 lakhs

Answer - Option D
Explanation -
Revenue of the sixth year = Revenue of the first 6 years + Revenue of the last 6 years – Revenue of total 11 years
= 6 × 64 + 6 × 76 – 11 × 69
= Rs 81 lakhs
3. The mean of marks secured by 55 students in division-A of class X is 55, 45 students of division-B is 51 and that of 40 students of division-C is 49. What is the mean of marks of the students of three divisions of Class X?

A. 51.3 B. 50.6 C. 52 D. 53.4

Answer - Option C
Explanation -
Mean of all the three divisions = [latex]\frac{sum of total marks secured by all the students}{total number of students}[/latex]
= [latex]\frac{55 \times 55 + 45 \times 51 + 40 \times 49}{55 + 45 + 40}[/latex]
= [latex]\frac{7280}{140}[/latex]
= 52
4. The sum of the ages of husband and wife at present is 100. Ten years ago the ratio of their ages was 9:7. What is the age of the husband?

A. 45 years B. 55 years C. 65 years D. 40 years

Answer - Option B
Explanation -
Let 10 years ago, the age of husband and wife be 9x and 7x respectively.
Given: (9x + 10) + (7y + 10) = 100
[latex]\rightarrow[/latex] 16x = 100 - 20
[latex]\rightarrow[/latex] x = 5
Thus, present age of husband = 9x + 10
= 9 × 5 + 10
= 55 years
5. The ratio of present ages of Rangana and Sayed is 7:5. After 11 years the ratio of their ages will be 4:3. What is Rangana's present age?

A. 55 B. 77 C. 70 D. 96

Answer - Option B
Explanation -
The ratio of present ages of Rangana and Sayed is 7:5. Let their present ages are 7x and 5x respectively. After 11 years, the ratio of their ages = 4:3, i.e.
[latex]\frac{7x + 11}{5x + 11}[/latex] = [latex]\frac{4}{3}[/latex]
21x + 33 = 20x + 44
x = 11
Thus, the present age of Rangana = 7x = 77 years

1. A is 4 years younger than B while C is 4 years younger than D but 1/5th times as old as A. If D is 8 years old, how many times as old is B as C?

A. 6 times B. 4 times C. 5 times D. 2 times

Answer - Option A
Explanation -
A = (B - 4),C = (D - 4)& C = [latex]\frac{1}{5}[/latex]A
C = 8 - 4 = 4
A = 5C = 5 * 4 = 20
B = A + 4 = 20 + 4 + 24
[latex]\frac{B}{C}[/latex] = [latex]\frac{24}{4}[/latex] = 6
B = 6C
2. If 10 years are subtracted from the present age of A and the remainder is divided by 6, then the present age of his grandson B is obtained. If B is 2 years younger to C whose age is 7 years, then what is A’s present age?

A. 40 years B. 55 years C. 52 years D. 65 years

Answer - Option A
Explanation -
[latex]\frac{A-10}{6}[/latex] = B, B = C - 2, C = 7 years
B = 7 - 2 = 5 years
[latex]\frac{A-10}{6}[/latex] =5
A = 40 years
3. The ratio of present ages of Ratnabali and Shaukat is 8:5. After 22 years the ratio of their ages will be 10:9. At present, what is Ratnabali’s age ?

A. 5 B. 14 C. 81 D. 8

Answer - Option D
Explanation -
Let the present age of Ratnabali= 8x
And present age of Shaukat= 5x
As per question
[latex]\frac{8x + 22}{5x + 22}[/latex] = [latex]\frac{10}{9}[/latex]
72x + 198 = 50x + 220
22x = 22
x = 1
hence present age of Ratnabali = 8× 1= 8 years
4. The ratio of present ages of Ranjini and Shahid is 5:4. After 13 years the ratio of their ages will be 6:5. What is Ranjini's present age?

A. 52 B. 65 C. 60 D. 32

Answer - Option B
Explanation -
Let present age of Rajini = 5x
And present age of Shahid = 4x
After 13 years,
Age of Rajini = 5x + 13
Age of Shahid = 4x + 13
Ratio of their age after 13 years is 6 : 5
[latex]\frac{(5x +13)}{(4x+13)}[/latex] = 6/5 [latex]\frac{6}{5}[/latex]
25x + 65 = 24x + 78
x = 13
Rajini’s present age = 5x = 5 × 13 = 65 years
5. The ratio of present ages of Rambha and Sarvesh is 8:5. After 7 years the ratio of their ages will be 5:4. What is Rambha's present age?

A. 5 B. 30 C. 8 D. 48

Answer - Option C
Explanation -
Let present age of Rambha and Sarvesh is 8x and 5x
After 7 years
[latex]\frac{8x + 7}{5x + 7} = \frac {5}{4}[/latex]
32x + 28 = 25x +35
7x = 7
X = 1
So Rambha present age will be 8x = 8 years

1. The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

A.28, 28 B. 32, 24 C. 30, 26 D. 29, 27

Answer - Option C
Explanation -
Let the present age of husband = x
And present age of wife = y
x + y = 56
x = 56 - y - - - - - - equation1
(x - 10)(y - 10) = 320
xy - 10x - 10y + 100 = 320
xy - 10(x + y) = 220
xy - 10(56) = 220
xy = 780 - - - - - - - equation 2
Put value of x from equation 1 into equation 2
(56 - y)y = 780
[latex]{y}^{2}[/latex] - 56y + 780 = 0
[latex]{y}^{2}[/latex] - 30y - 26y + 780 = 0
y = 30 or 26
so we get x = 26 or 30
hence the required ages are 30 and 26
2. A shopkeeper, sold dried apricots at the rate Rs 1210 a kg and bears a loss of 12%. Now if he decides to sell it at Rs 1331 per kg, what will be the result?

A. 6.4 percent loss B. 3.2 percent gain C. 6.4 percent gain D. 3.2 percent loss

Answer - Option D
Explanation -
Old selling price = 1210 and loss percentage = 12%
Loss percentage = [latex]\frac{CP - SP}{CP}* 100[/latex]
This means (100 - 12)% of CP = 1210
88% CP = 1210 * [latex]\frac{100}{88}[/latex]
CP = 1210
CP = 1375
New selling price = 1331
Loss = CP-SP = 1375 - 1331 = 44
Loss percentage = [latex]44 * \frac{100}{1375}[/latex]
loss percentage = 3.2 %
3. [latex]\frac{4}{5}[/latex] part of a tank is filled with oil. After taking out 42 litres of oil the tank is 3/4 part full. What is the capacity (in litres) of the tank?

A. 420 B. 630 C. 840 D. 1680

Answer - Option C
Explanation -
Let the capacity of tank be 20x
Then initial quantity of oil in the tank= [latex]\frac{4}{5}[/latex] * 20X = 16X
Final quantity = [latex]\frac{3}{4}[/latex] * 20X = 15X
16x - 15x = 42
X = 42
So capacity of tank=20x = 20 × 42 = 840
4. To do a certain work, B would take time thrice as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. The time taken by A to complete the work separately is

A. 22 days B. 24 days C. 30 days D. 20 days

Answer - Option B
Explanation -
If B does the work in 3x days, (A + C) will do the same work in x days.
If C does that work in 2y days. (A + B) will do it in y days.
therefore, [latex]\frac{1}{x}\frac{1}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{4}{3x}[/latex] = [latex]\frac{1}{10}[/latex]
3x = 40
x = [latex]\frac{40}{3}[/latex]
Again, [latex]\frac{1}{y}\frac{1}{2y}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{3}{2y}[/latex] = [latex]\frac{1}{10}[/latex] = y = 15
therefore(A + B+ C)'S = 1 Day's work = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A} + \frac{1}{40} + \frac{1}{30}[/latex] = [latex]\frac{1}{10}[/latex]
[latex]\frac{1}{A}[/latex] = [latex]\frac{1}{10} - \frac{1}{40} - \frac{1}{30}[/latex]
[latex]\frac{12 - 3- 4}{120}[/latex] = [latex]\frac{5}{120}[/latex] = [latex]\frac{1}{24}[/latex]
therefore, A alone will complete the work in 24 days.
5. The sum of the ages of husband and wife at present is 100. Ten years ago the ratio of their ages was 9:7. What is the age of the husband?

A. 45 years B. 55 years C. 65 years D. 40 years

Answer - Option B
Explanation -
Let 10 years ago, the age of husband and wife be 9x and 7x respectively.
Given: (9x + 10) + (7y + 10) = 100
[latex]\rightarrow[/latex] 16x = 100 - 20
[latex]\rightarrow[/latex] x = 5
Thus, present age of husband = 9x + 10
= 9 × 5 + 10
= 55 years

Direction(1-5): Study the following pie chart carefully to answer the following questions given below

No of Students : 5,00,000

1. What is the ratio of no of boys in Economy to no of girls in Hotel Management ?

A. 47:81 B. 168:125 C. 119:92 D. 17:40

Answer - Option B
Explanation -
Economy = 12% of 500000 = 60,000
Boys in Economy = 56% of 60,000 = 33600
HM =25% of 500000 = 1,25,000
Girls in HM = 20% of 1,25,000 = 25000
336:250 = 168:125
2. What is the difference between the no of girls in IT and no of girls in ECE ?

A. 2100 B. 5600 C. 3200 D. 4800

Answer - Option D
Explanation -
No of student in ECE = 16 % of 5,00,000 = 80,000
Girls in ECE = 29% of 80,000 = 23200
No of Students in IT = 14 % of 5,00,000 = 70,000
Girls in IT = 40% of 70,000 = 28000
Difference = 28000 – 23200 = 4800
3. The number of boys in CS is approximately what % of total number of students in EEE ?

A. 36% B. 41% C. 27% D. 30%

Answer - Option A
Explanation -
No of students in CS = 15% of 5,00,000 = 75000
Boys in CS = 43% of 75000 = 32250
No of Students in EEE = 18% of 5,00,000 = 90,000
% of Boys = 32250*100/90,000 = 35.8 = 36%
4. What is the average number of girls students in the fields ECE, EEE and Hotel Management ?

A. 32,560 B. 56,260 C. 26,566 D. 41,256

Answer - Option C
Explanation -
Girls in ECE = 23,200
Girls in EEE = 31,500
Girls in HM = 25,000
Average = [latex]\frac {79700}{3}[/latex] = 26,566.66 = 26,566
5. The large number of girls students distributed in which field ?

A. ECE B. CS C. Hotel Management D. IT

Answer - Option B
Explanation -
Girls in ECE = 23,200
Girls in EEE = 31,500
Girls in HM = 25,000
Girls in CS = 42,750
Girls in IT = 28,000
Girls in Economy = 26,400