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3889 + 12.952 - ? = 3854.002- a)47.095
- b)47.752
- c)47.932
- d)47.95
Correct answer is option 'D'. Can you explain this answer?
3889 + 12.952 - ? = 3854.002
a)
47.095
b)
47.752
c)
47.932
d)
47.95
|
Ishaan Roy answered |
Calculation:
Given: 3889 + 12.952 - ? = 3854.002
Find the missing number:
To find the missing number, we need to solve the equation by rearranging the terms.
Step 1:
Add 3889 and 12.952:
3889 + 12.952 = 3901.952
Step 2:
Substitute the sum back into the equation:
3901.952 - ? = 3854.002
Step 3:
Now, subtract 3854.002 from 3901.952:
3901.952 - 3854.002 = 47.95
Therefore, the missing number is 47.95.
Answer:
So, the correct answer is option 'D' 47.95.
Given: 3889 + 12.952 - ? = 3854.002
Find the missing number:
To find the missing number, we need to solve the equation by rearranging the terms.
Step 1:
Add 3889 and 12.952:
3889 + 12.952 = 3901.952
Step 2:
Substitute the sum back into the equation:
3901.952 - ? = 3854.002
Step 3:
Now, subtract 3854.002 from 3901.952:
3901.952 - 3854.002 = 47.95
Therefore, the missing number is 47.95.
Answer:
So, the correct answer is option 'D' 47.95.
If 1 is added to the denominator of a fraction it becomes 1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is- a)14
- b)12
- c)10
- d)6
Correct answer is option 'D'. Can you explain this answer?
If 1 is added to the denominator of a fraction it becomes 1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is
a)
14
b)
12
c)
10
d)
6
|
Target Study Academy answered |
Let the numerator = x and denominator = y
y + 3 = 2y ⇒ y = 3
x +1= 3 ⇒ x = 2
∴ xy = 2 × 3 = 6
The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was ₹ 4.20 and that of Y was ₹ 6.30, in which year commodity X will cost 40 paise more than the commodity?
- a)2012
- b)2011
- c)2013
- d)2010
Correct answer is option 'B'. Can you explain this answer?
The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was ₹ 4.20 and that of Y was ₹ 6.30, in which year commodity X will cost 40 paise more than the commodity?
a)
2012
b)
2011
c)
2013
d)
2010
|
Malavika Rane answered |
Given information:
- The price of commodity X increases by 40 paise every year.
- The price of commodity Y increases by 15 paise every year.
- In 2001, the price of commodity X was 4.20 and that of Y was 6.30.
Objective:
To determine the year in which commodity X will cost 40 paise more than commodity Y.
1. Calculate the difference in prices between commodity X and Y in 2001.
2. Determine the number of years required for commodity X to cost 40 paise more than commodity Y.
3. Add the calculated number of years to 2001 to find the year in which commodity X will cost 40 paise more than commodity Y.
The difference in prices between commodity X and Y in 2001: 6.30 - 4.20 = 2.10
Let's assume 'n' is the number of years required for commodity X to cost 40 paise more than commodity Y.
Price increase for X in 'n' years: 40n paise
Price increase for Y in 'n' years: 15n paise
According to the given information, the difference in prices between commodity X and Y in 'n' years will be 40 paise.
Equation: 2.10 + (40n - 15n) = 40
Simplifying the equation: 2.10 + 25n = 40
25n = 40 - 2.10
25n = 37.90
n = 37.90 / 25
n ≈ 1.516
Therefore, it will take approximately 1.516 years for commodity X to cost 40 paise more than commodity Y.
Adding 1.516 years to 2001, we get:
2001 + 1.516 = 2002.516
Since the question asks for the year, we round up the decimal value to the nearest year.
Therefore, commodity X will cost 40 paise more than commodity Y in the year 2003.
Conclusion:
Commodity X will cost 40 paise more than commodity Y in the year 2003.
- The price of commodity X increases by 40 paise every year.
- The price of commodity Y increases by 15 paise every year.
- In 2001, the price of commodity X was 4.20 and that of Y was 6.30.
Objective:
To determine the year in which commodity X will cost 40 paise more than commodity Y.
Approach:
1. Calculate the difference in prices between commodity X and Y in 2001.
2. Determine the number of years required for commodity X to cost 40 paise more than commodity Y.
3. Add the calculated number of years to 2001 to find the year in which commodity X will cost 40 paise more than commodity Y.
Calculation:
The difference in prices between commodity X and Y in 2001: 6.30 - 4.20 = 2.10
Let's assume 'n' is the number of years required for commodity X to cost 40 paise more than commodity Y.
Price increase for X in 'n' years: 40n paise
Price increase for Y in 'n' years: 15n paise
According to the given information, the difference in prices between commodity X and Y in 'n' years will be 40 paise.
Equation: 2.10 + (40n - 15n) = 40
Simplifying the equation: 2.10 + 25n = 40
25n = 40 - 2.10
25n = 37.90
n = 37.90 / 25
n ≈ 1.516
Therefore, it will take approximately 1.516 years for commodity X to cost 40 paise more than commodity Y.
Year:
Adding 1.516 years to 2001, we get:
2001 + 1.516 = 2002.516
Since the question asks for the year, we round up the decimal value to the nearest year.
Therefore, commodity X will cost 40 paise more than commodity Y in the year 2003.
Conclusion:
Commodity X will cost 40 paise more than commodity Y in the year 2003.
A number exceeds its one-fifth by 20. The number is- a)5
- b)20
- c)25
- d)100
Correct answer is option 'C'. Can you explain this answer?
A number exceeds its one-fifth by 20. The number is
a)
5
b)
20
c)
25
d)
100
|
|
Malavika Rane answered |
Understanding the Problem
In this problem, we need to find a number that exceeds its one-fifth by 20. Let's denote the unknown number as "x".
Setting Up the Equation
1. Identify One-Fifth of the Number:
One-fifth of the number can be expressed as:
- (1/5)x = x/5
2. Formulate the Condition:
According to the problem, the number exceeds its one-fifth by 20, which can be written as:
- x = (x/5) + 20
Solving the Equation
1. Multiply the Entire Equation by 5:
To eliminate the fraction, multiply every term by 5:
- 5x = x + 100
2. Rearrange the Equation:
Subtract x from both sides:
- 5x - x = 100
- 4x = 100
3. Divide by 4:
Now, divide both sides by 4 to isolate x:
- x = 100 / 4
- x = 25
Conclusion
The calculated value of x is 25. Therefore, the number that exceeds its one-fifth by 20 is 25, which corresponds to option 'C'.
Verification
- One-fifth of 25 is 5.
- 25 exceeds 5 by 20, confirming our solution.
In this problem, we need to find a number that exceeds its one-fifth by 20. Let's denote the unknown number as "x".
Setting Up the Equation
1. Identify One-Fifth of the Number:
One-fifth of the number can be expressed as:
- (1/5)x = x/5
2. Formulate the Condition:
According to the problem, the number exceeds its one-fifth by 20, which can be written as:
- x = (x/5) + 20
Solving the Equation
1. Multiply the Entire Equation by 5:
To eliminate the fraction, multiply every term by 5:
- 5x = x + 100
2. Rearrange the Equation:
Subtract x from both sides:
- 5x - x = 100
- 4x = 100
3. Divide by 4:
Now, divide both sides by 4 to isolate x:
- x = 100 / 4
- x = 25
Conclusion
The calculated value of x is 25. Therefore, the number that exceeds its one-fifth by 20 is 25, which corresponds to option 'C'.
Verification
- One-fifth of 25 is 5.
- 25 exceeds 5 by 20, confirming our solution.
How many digits will be there to the right of the decimal point in the product of 95.75 and .02554?- a)5
- b)6
- c)7
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
How many digits will be there to the right of the decimal point in the product of 95.75 and .02554?
a)
5
b)
6
c)
7
d)
None of these
|
EduRev SSC CGL answered |
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.
The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction is increased by 1/24. The difference of numerator and denominator of the fraction is- a)9
- b)1
- c)3
- d)5
Correct answer is option 'B'. Can you explain this answer?
The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction is increased by 1/24. The difference of numerator and denominator of the fraction is
a)
9
b)
1
c)
3
d)
5
|
|
EduRev SSC CGL answered |
Let numerator be x, then denominator = 11 – x.
⇒ 528 – 96x = 143 – 24x + x2
⇒ x2 + 72x – 385 = 0
⇒ x2 + 77x – 5x – 385 = 0
⇒ x(x + 77) − 5(x + 77) = 0
⇒ (x - 5) (x + 77) = 0 ⇒ x = 5
Denominator = 11 - 5 = 6
Difference = 6 - 5 = 1
in decimal for is:
A candidate in an examination was asked to find 
of a certain number. By mistake he found 
of it. Thus, his answer was 25 more than the correct answer. The number was :- a)140
- b)84
- c)56
- d)28
Correct answer is option 'D'. Can you explain this answer?
A candidate in an examination was asked to find of a certain number. By mistake he found of it. Thus, his answer was 25 more than the correct answer. The number was :
of a certain number. By mistake he found of it. Thus, his answer was 25 more than the correct answer. The number was :a)
140
b)
84
c)
56
d)
28
|
|
EduRev SSC CGL answered |
Let the required number be x.
As given,
As given,
0.1 and 5/8 of a bamboo are in mud and water respectively and the rest of length 2.75 m is above water. What is the length of the bamboo?- a)20 m
- b)27.5 m
- c)30 m
- d)10 m
Correct answer is option 'D'. Can you explain this answer?
0.1 and 5/8 of a bamboo are in mud and water respectively and the rest of length 2.75 m is above water. What is the length of the bamboo?
a)
20 m
b)
27.5 m
c)
30 m
d)
10 m
|
Ssc Cgl answered |
Let the length of bamboo be x metres.
Length of bamboo above water
Length of bamboo above water
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