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Contents:

- Aptitude
- Approximation
- Average
- Boat and Stream
- Compound interest
- Discount
- Linear Equations
- Mensuration
- Mixture and Allegation
- Number series
- Number System
- Partnership
- Percentage
- Permutation and combination
- Pipes and Cisterns
- Probability
- Problem on ages
- Profit and Loss
- Ratio and Proportions
- Simple and compound interest
- Time and Distance
- Time and work
- Trains
- Unitary Method
- Word problems
- Work and Wages

11). The sum of five consecutive odd number is equal to 175. What is the sum of the second largest number and the square of the smallest number amongst them together?
Sum of five consecutive odd numbers = x + x + 2 + x + 4 + x + 6 + x + 8 = 175 = 5x + 20 = 175 = \( \Large x = \frac{175 - 20}{5} = 31 \) Sum of the second largest and square smallest one =(31+ 6) + (31)2 = 37 + 961 = 998 = \( \Large \left(31 + 6\right) + \left(31\right)^{2} \) = 37 + 961 = 998 | ||||

12). 0ut of three given numbers, the first number is twice the second and thrice the third. If the average of these three numbers is 154, then what is the difference between the first and the third numbers?
Let the third number = x Then, the first number = 3x and second number = \( \Large \frac{3x}{2} \) According to the question, \( \Large \frac{x + 3x + \frac{3x}{2}}{3} = 154 \) \( \Large \frac{2x + 6x + 3x}{6} = 154 \) \( \Large x = \frac{154 \times 6}{11} = 84 \) Therefore, Required difference = 3x - x = 2x \( \Large 2 \times 84 = 168 \) | ||||

13). X, Y and Z had taken a dinner together. The cost of the meal of Z was 20% more than that of Y and the cost of the meal of X was \( \Large \frac{5}{6} \) as much as the cost of the meal of Z. If Y paid Rs.100, then what was the total amount that all the three of them had paid?
Given that, The cost of meal of y = Rs.100 Now, according to the question, The cost of the meal of z = 20% more than that of y \( \Large \left(100 + \frac{20}{100} \times 100\right)= \left(100 + 20\right) = Rs.120 \) and the cost of the meal of x = \( \Large \frac{5}{6} \) as much as the cost of the meal of z \( \Large \frac{5}{6} \times 120 \) = Rs.100 Therefore, Total amount that all the three of them has paid = 100 + 120 + 100 = Rs.320 | ||||

14). Find the maximum number of trees which can be planted 20 m apart on the two sides of a straight road 1760 m long.
Number of trees that can be planted on one side of road = \( \Large \frac{1760}{20} + 1 \) = 88 + 1 = 89 Trees on the both sides = \( \Large 2 \times 89 = 178 \) | ||||

15). The difference between two numbers is 18. If four times the second number is less than three times the first number by 18, then what is the sum of these two numbers?
Let first number = x and second number = y According to the question. x - y = 18 ...(i) and 3x - 4y = 18 ...(ii) On multiplying Eq. (i) by 3 and then subtracting Eq. (ii) from it, we get 3x - 3y = 54 3x - 4y = 18 y = 36 On putting the value of y in Eq. (i), we get x = 18 + y = 18 + 36 => x = 54 Therefore, Required sum = x + y = 54+ 36 = 90 | ||||

16). The difference between a two-digit number and the number obtained by interchanging the two digits of the number is 18. The sum of the two digits of the number is 12. What is the product of the digits of two-digit number?
Let the unit's digit be y and ten's digit, be x. | ||||

17). On Children's Day, sweets were to be equally distributed amongst 300 children. But on that particular day 50 children remained absent; hence each child got one extra sweet. How many sweets were distributed?
Let total number of sweets = x | ||||

18). There are two examination halls P and Q. If 10 students shifted P to Q, then the number of students will be equal in both the examination halls. If 20 students shifted from Q to P, then the students of P would be doubled to the students of Q. The numbers of students would be in P and Q, respectively are
Let number of students in examination halls P and Q is x and y, respectively. Then, as per the first condition, x - 10 = y + 10 = x - y = 20 ...(i) As per the second condition, \( \Large x + 20 = 2 \left(y - 20\right) \) => x + 20 = 2y - 40 => x - 2y = -60 ...(ii) On subtracting Eq. (ii) from Eq. (i), we get -y + 2y = 20 + 60 => y = 80 Putting the value of y in Eq. (i), we get x - 80 = 20 => x =100 Hence, number of students in examination halls P and Q is 100 and 80, respectively. | ||||

19). In a two-digit positive number, the unit digit is equal to the square of ten's digit. The difference between the original number and the number formed by interchanging the digits is 54. What is 40% of the original number?
Let ten's digit be x and unit's digit be x2. | ||||

20). In a three-digit number, the digit in the unit's place is four times the digit in the hundred's place. If the digit in the unit's place and the ten's place are interchanged, the new number so formed is 18 more than the original number. If the digit in the hundreds place is one-third of the digit in the ten's place, then what is 25% of the original number?
Let hundred's digit = x Then unit's digit = 4x and ten's digit = 3x Number = 100x + 30x + 4x = 134x Again, hundred's digit = x Ten's digit = 4x and unit's digit = 3x Number = 100x + 40x + 3x = 143x According to the question, 143x -134X = 18 => 9x = 18 x = 2 Original number = 134x = 134 x 2 = 268 25% of original number = \( \Large 268 \times \frac{25}{100} = 67 \) |