Number System
- A and B have together three times what B and C have, while A, B, C together have thirty rupees more than that of A. If B has 5 times that of C, then A has
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A + B= 3 (B + C)
A + B + C = A + 30
B = 5 C
∴ A + B = 3 (B + C)
⇒ A + 5C = 18C ⇒ A = 13C
∴ A + B + C = A + 30
13C + 5C + C = 13C + 30
⇒ 6C = 30
⇒ C = 5
⇒ A = 13 × 5 = ₹ 65Correct Option: B
A + B= 3 (B + C)
A + B + C = A + 30
B = 5 C
∴ A + B = 3 (B + C)
⇒ A + 5C = 18C ⇒ A = 13C
∴ A + B + C = A + 30
13C + 5C + C = 13C + 30
⇒ 6C = 30
⇒ C = 5
⇒ A = 13 × 5 = ₹ 65
- Find the maximum number of trees which can be planted, 20 metres apart, on the two sides of a straight road 1760 metres long
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Number of trees on each side of the road
[+1 because we would start with a tree]= 1760 + 1 = 88 + 1 = 89 20
∴ Required answer
= 89 × 2 = 178Correct Option: B
Number of trees on each side of the road
[+1 because we would start with a tree]= 1760 + 1 = 88 + 1 = 89 20
∴ Required answer
= 89 × 2 = 178
- If
then the value of 12 * 4 is :a * b = a + b + a b
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a * b = a + b + a b ∴ 12 * 4 = 12 + 4 + 12 4
= 16 + 3 = 19Correct Option: D
a * b = a + b + a b ∴ 12 * 4 = 12 + 4 + 12 4
= 16 + 3 = 19
- Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is :
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If the number of correct sums be x, then,
x × 3 – (30 –x) × 2 = 40
⇒ 3x – 60 + 2x = 40
⇒ 5x = 60 + 40 = 100
⇒ x = 20Correct Option: B
If the number of correct sums be x, then,
x × 3 – (30 –x) × 2 = 40
⇒ 3x – 60 + 2x = 40
⇒ 5x = 60 + 40 = 100
⇒ x = 20
- The product of two positive numbers is 2500. If one number is four times the other, the sum of the two numbers is :
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Let one of the positive number be x.
∴ The other will be 4x
Now, 4x × x = 2500
⇒ x2 = 2500 ÷ 4 = 625
∴ x = √625 = 25
∴ Sum of the two numbers
4x + x = 5x = 5 × 25 =125Correct Option: B
Let one of the positive number be x.
∴ The other will be 4x
Now, 4x × x = 2500
⇒ x2 = 2500 ÷ 4 = 625
∴ x = √625 = 25
∴ Sum of the two numbers
4x + x = 5x = 5 × 25 =125
