Question Detail

if the sum of $\begin{aligned} \frac{1}{2} \end{aligned}$ and $\begin{aligned} \frac{1}{5} \end{aligned}$ of a number exceeds $\begin{aligned} \frac{1}{3} \end{aligned}$ of the number by $\begin{aligned} 7\frac {1}{3} \end{aligned}$ , then number is

Answer: Option B

Explanation:

Seems a bit complicated, isnt'it, but trust me if we think on this question with a cool mind then it is quite simple...
Let the number is x,
then,

$\begin{aligned} (\frac{1}{2}x + \frac{1}{5}x) - \frac{1}{3}x = \frac{22}{3} \end{aligned}$

$\begin{aligned} => \frac{11x}{30} = \frac{22}{3} \end{aligned}$

$\begin{aligned} => x = 20 \end{aligned}$

1. Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

Answer: Option A

Explanation:

Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5

2. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

Answer: Option A

Explanation:

=> 3(2x+9) = 75
=> 2x+9 = 25
=> x = 8

3. Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

10:3
5:9
11:9
13:9

Answer: Option C

Explanation:

$\begin{aligned} => \frac{(x+y)}{(x-y)} = \frac{40}{4} \end{aligned}$

$\begin{aligned} => (x+y)= 10(x-y) \end{aligned}$

$\begin{aligned} => 9x = 11y => \frac{x}{y} = \frac{11}{9} \end{aligned}$

4. Sum of two numbers is 25 and their difference is 13. Find their product.

Answer: Option C

Explanation:

Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:

$\begin{aligned} => (x+y)^2 = (x-y)^2 + 4xy \end{aligned}$

$\begin{aligned} <=> (25)^2 = (13)^2 + 4xy \end{aligned}$

$\begin{aligned} <=> 4xy = (25)^2 - (13)^2 \end{aligned}$

$\begin{aligned} <=> xy = \frac{456}{4} = 114 \end{aligned}$

5. Two numbers differ by 5. If their product is 336, then sum of two number is

Answer: Option D

Explanation:

Friends you remember,

$\begin{aligned} => (x+y)^2 = (x-y)^2 + 4xy \end{aligned}$

$\begin{aligned} => (x+y)^2 = (5)^2 + 4(336) \end{aligned}$

$\begin{aligned} => (x+y) = \sqrt{1369} = 37 \end{aligned}$

Question Detail

if the sum of $\begin{aligned} \frac{1}{2} \end{aligned}$ and $\begin{aligned} \frac{1}{5} \end{aligned}$ of a number exceeds $\begin{aligned} \frac{1}{3} \end{aligned}$ of the number by $\begin{aligned} 7\frac {1}{3} \end{aligned}$ , then number is

1. Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

2. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

3. Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

4. Sum of two numbers is 25 and their difference is 13. Find their product.

5. Two numbers differ by 5. If their product is 336, then sum of two number is

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Question Detail

if the sum of 12\begin{aligned} \frac{1}{2} \end{aligned} and 15\begin{aligned} \frac{1}{5} \end{aligned} of a number exceeds 13\begin{aligned} \frac{1}{3} \end{aligned} of the number by 713\begin{aligned} 7\frac {1}{3} \end{aligned}, then number is

1. Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

2. A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number

3. Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

4. Sum of two numbers is 25 and their difference is 13. Find their product.

5. Two numbers differ by 5. If their product is 336, then sum of two number is

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if the sum of $\begin{aligned} \frac{1}{2} \end{aligned}$ and $\begin{aligned} \frac{1}{5} \end{aligned}$ of a number exceeds $\begin{aligned} \frac{1}{3} \end{aligned}$ of the number by $\begin{aligned} 7\frac {1}{3} \end{aligned}$ , then number is