Simplification
- One-third of Rahul's marks in Mathematics exceeds one-half of his marks in Hindi by 30. If he got 480 marks in the two subjects together, how many marks did he get in Hindi ?
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Let Rahul's marks in Mathematics = x
and in Hindi = y
According to the question,
1/3 x - 1/2y = 30
⇒ 2x - 3y = 180 ...(i)
Also, given
x + y = 480 ...(ii)
Solve the above equation.Correct Option: C
Let Rahul's marks in Mathematics = x
and in Hindi = y
According to the question,
1/3 x - 1/2y = 30
⇒ 2x - 3y = 180 ...(i)
Also, given
x + y = 480 ...(ii)
By solving Eqs. (i) and (ii), we get
x = 324
and y =156
- A container of milk was 4/5 full. When 12 bottles of milk were taken out and 8 bottles of milk were poured into it, it was 3/4 full. How many bottles of milk can the container contain ?
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Let x bottles can fill the container completely.
According to the questions,
4/5 x - 3/4 x = (12 - 8)Correct Option: A
Let x bottles can fill the container completely.
According to the questions,
4/5 x - 3/4 x = (12 - 8)
⇒ 16x - 15x/20 = (12 - 8)
⇒ x/20 = 4
∴ x = 80
∴ Required number of bottles = 80.
- The number obtained by interchanging the digits of a two-digit number is more than the original number by 27 and the sum of the digits is 13. What is the original number ?
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Let the ten's place digit = x
and unit's place digit = y
∴ The original number = 10x + y
After interchanging the digits
Number will become 10y + x
∴ according to the given question,
y + x = 13 and
(10y + x ) - (10x + y) = 27Correct Option: A
Let the ten's place digit = x
and unit's place digit = y
∴ The original number = 10x + y
After interchanging the digits
Number will become 10y + x
∴ according to the given question,
y + x = 13 and
(10y + x ) - (10x + y) = 27
⇒ 9y - 9x = 27
⇒ y - x = 3 ................(i)
and y + x = 13 .................(ii)
On solving Eqs.(i) and (ii) , we get
y = 8 and x = 5
∴ Required number is
10x + y = 10 x 5 + 8 = 58
- Of three positive numbers the product of the first and second is 42, that of the second and third is 56 and that of third and first is 48, The third number is
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Let first, second and third numbers be x , y and z, respectively.
Then , xy = 42.................(i)
yz = 56 ..................(ii)
xz = 48 ....................(iii)
Multiplying Eqs ( i), (ii) and (iii), we get
(xyz)2 = 42 x 56 x 48
⇒ (xyz)2 = 112896
⇒ xyz = 336 ................(iv)
Dividing Eq. (iv) by Eq. (i), we get
xyz/xy = 336/42 ⇒ z = 8Correct Option: D
Let first, second and third numbers be x , y and z, respectively.
Then , xy = 42.................(i)
yz = 56 ..................(ii)
xz = 48 ....................(iii)
Multiplying Eqs ( i), (ii) and (iii), we get
(xyz)2 = 42 x 56 x 48
⇒ (xyz)2 = 112896
⇒ xyz = 336 ................(iv)
Dividing Eq. (iv) by Eq. (i), we get
xyz/xy = 336/42 ⇒ z = 8
- In a certain office (1/3) of the worker are women (1/2) of the women are married and (1/3) of the married women have children. If (3/4) of the man are married and (2/3) of the married men have children. What part of workers are without children ?
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Let, total number of workers be N .
Then number of women = N/3
Number of men = 2N/3
Number of women having children = 1/3 of 1/2 of N/3 = N/18
Number of men having children = 2/3 of 3/4 of 2N/3 = N/3
Number of workers having children = N/18 + N/3 = 7N/18
Number of workers having no children = N - 7N/18 = 11N/18Correct Option: C
Let, total number of workers be N .
Then number of women = N/3
Number of men = 2N/3
Number of women having children = 1/3 of 1/2 of N/3 = N/18
Number of men having children = 2/3 of 3/4 of 2N/3 = N/3
Number of workers having children = N/18 + N/3 = 7N/18
Number of workers having no children = N - 7N/18 = 11N/18
=11/18 of all workers
