Simplification
- The difference between two numbers is 9 and the difference between their squares is 207. The numbers are :
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Let the numbers be a and b where a > b.
According to the question,
a – b = 9 .... (i)
and a2 – b2 = 207
⇒ (a + b) (a – b) = 207
⇒ 9 (a + b) = 207⇒ a + b = 207 = 23 .... (ii) 9
On adding equations (i) and (ii),
a + b + a – b = 23 + 9
⇒ 2a = 32 ⇒ a = 16
∴ a – b = 9
⇒ 16 – b = 9
⇒ b = 16 – 9 = 7Correct Option: B
Let the numbers be a and b where a > b.
According to the question,
a – b = 9 .... (i)
and a2 – b2 = 207
⇒ (a + b) (a – b) = 207
⇒ 9 (a + b) = 207⇒ a + b = 207 = 23 .... (ii) 9
On adding equations (i) and (ii),
a + b + a – b = 23 + 9
⇒ 2a = 32 ⇒ a = 16
∴ a – b = 9
⇒ 16 – b = 9
⇒ b = 16 – 9 = 7
- The sum of squares of three positive integers is 323. If the sum of squares of two numbers is twice the third, their product is
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Let the positive numbers be x, y and z (respectively).
∴ x2 + y2 + z2 = 323 ... (i)
and x2 + y2 = 2z ... (ii)
∴ z2 + 2z = 323
⇒ z2 2z – 323 = 0
⇒ z2 + 19z – 17z – 323 = 0
⇒ z (z + 19) – 17 (z + 19) = 0
⇒ (z – 17) (z + 19) = 0
⇒ z = 17 because z ≠ –19
∴ x2 + y2 = 2 × 17 = 34
= 32 + 52
∴ xyz = 3 × 5 × 17 = 255Correct Option: A
Let the positive numbers be x, y and z (respectively).
∴ x2 + y2 + z2 = 323 ... (i)
and x2 + y2 = 2z ... (ii)
∴ z2 + 2z = 323
⇒ z2 2z – 323 = 0
⇒ z2 + 19z – 17z – 323 = 0
⇒ z (z + 19) – 17 (z + 19) = 0
⇒ (z – 17) (z + 19) = 0
⇒ z = 17 because z ≠ –19
∴ x2 + y2 = 2 × 17 = 34
= 32 + 52
∴ xyz = 3 × 5 × 17 = 255
- Each member of a club contributes as much rupees and as much paise as the number of members of the club. If the total contribution is Rs. 2525, then the number of members of the club is
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Number of members in the club = x (let)
According to the question,x2 + x2 = 2525 100 ⇒ 100x2 + x2 = 2525 100
⇒ 101x2 = 252500⇒ x2 = 252500 = 2500 101
⇒ x = √2500 = 50Correct Option: D
Number of members in the club = x (let)
According to the question,x2 + x2 = 2525 100 ⇒ 100x2 + x2 = 2525 100
⇒ 101x2 = 252500⇒ x2 = 252500 = 2500 101
⇒ x = √2500 = 50
- A number when divided by 6 leaves remainder 3. When the square of the same number is divided by 6, the remainder is:
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Let the number (n) be 6m + 3
where m = quotient.
On squaring both sides,
n2 = (6m + 3)2
= 36 m2 + 36 m + 32
∴ Required remainder = 32
∴ Remainder on dividing 9 by 6 = 3Correct Option: D
Let the number (n) be 6m + 3
where m = quotient.
On squaring both sides,
n2 = (6m + 3)2
= 36 m2 + 36 m + 32
∴ Required remainder = 32
∴ Remainder on dividing 9 by 6 = 3
- The product of two positive integers is 2048 and one of them is twice the other. The smaller number is
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Let the smaller number be x.
∴ Larger number = 2x
According to the question,
2x2 = 2048⇒ x2 = 2048 = 1024 2
∴ x = √1024 = 32Correct Option: A
Let the smaller number be x.
∴ Larger number = 2x
According to the question,
2x2 = 2048⇒ x2 = 2048 = 1024 2
∴ x = √1024 = 32
