Simplification
- Find the least number which must be subtracted from 18265 to make it a perfect square.
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∴ Required answer = 40Correct Option: C
∴ Required answer = 40
- If a perfect square, not divisible by 6, be divided by 6, the remainder will be
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Remainder on dividing 32 = 9 by 6 = 3
Remainder on dividing 42 = 16 by 6 = 4
Remainder on dividing 52 = 25 by 6 = 1Correct Option: C
Remainder on dividing 32 = 9 by 6 = 3
Remainder on dividing 42 = 16 by 6 = 4
Remainder on dividing 52 = 25 by 6 = 1
- The greatest perfect square number of 6 digits is
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Largest 6-digit number = 999999
∴ Required perfect square number = 999999 – 1998 = 998001Correct Option: B
Largest 6-digit number = 999999
∴ Required perfect square number = 999999 – 1998 = 998001
- three numbers are such that their sum is 50, product is 3750 and the sum of their reciprocals is 31/50 .Find the sum of the squares of the three numbers.
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x + y + z = 50 ; xyz = 3750
∴ 1 + 1 + 1 = yz + zx + xy x y z xyz = 31 150 ⇒ xy + yz + zx = 31 xyz 150 = 31 × 3750 = 775 150
∴ (x + y + z )2 = x2 + y2 + z2 + 2(xy + yz + zx)
⇒ (50)2 + x2 + y2 + z2 + 2 × 775
⇒ 2500 = x2 + y2 + z2 + 1550
⇒ x2 + y2 + z2 = 2500 – 1550
= 950Correct Option: C
x + y + z = 50 ; xyz = 3750
∴ 1 + 1 + 1 = yz + zx + xy x y z xyz = 31 150 ⇒ xy + yz + zx = 31 xyz 150 = 31 × 3750 = 775 150
∴ (x + y + z )2 = x2 + y2 + z2 + 2(xy + yz + zx)
⇒ (50)2 + x2 + y2 + z2 + 2 × 775
⇒ 2500 = x2 + y2 + z2 + 1550
⇒ x2 + y2 + z2 = 2500 – 1550
= 950
- The sum of three positive numbers is 18 and their product is 162. If the sum of two numbers is equal to the third number, then the sum of squares of the numbers is
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Let three positive integers be x, y and z.
According to the question,
x + y + z = 18 ..... (i)
xyz = 162 ..... (ii)
and x + y = z ..... (iii)
From equation (i),
z + z = 18 ⇒ 2z = 18 ⇒ z = 9
∴ xyz = 162
⇒ xy × 9 = 162⇒ xy = 162 = 18 ..... (iv) 9
∴ (x – y)2 = (x + y)2 – 4xy
= (9)2 – 4 × 18
= 81 – 72 = 9
∴ x – y = 3
∴ x + y + x – y = 9 + 3
⇒ 2x = 12 ⇒ x = 6
∴ x + y + z = 18
⇒ 6 + y + 9 = 18
⇒ y = 18 – 15 = 3
∴ x2 + y2 + z2
= (6)2 + (3)2 + (9)2
= 36 + 9 + 81 = 126Correct Option: B
Let three positive integers be x, y and z.
According to the question,
x + y + z = 18 ..... (i)
xyz = 162 ..... (ii)
and x + y = z ..... (iii)
From equation (i),
z + z = 18 ⇒ 2z = 18 ⇒ z = 9
∴ xyz = 162
⇒ xy × 9 = 162⇒ xy = 162 = 18 ..... (iv) 9
∴ (x – y)2 = (x + y)2 – 4xy
= (9)2 – 4 × 18
= 81 – 72 = 9
∴ x – y = 3
∴ x + y + x – y = 9 + 3
⇒ 2x = 12 ⇒ x = 6
∴ x + y + z = 18
⇒ 6 + y + 9 = 18
⇒ y = 18 – 15 = 3
∴ x2 + y2 + z2
= (6)2 + (3)2 + (9)2
= 36 + 9 + 81 = 126
