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Let the unit digit be 'a' and tens digit be 'b'.
Original number = 10b + a
Reversed number = 10a + b
Given:
a + b = 9 --------- (I)
10b + a - 27 = 10a + b
⇒ 9b - 9a = 27
⇒ b - a = 3 --------- (II)
Add equations (I) and (II):
a + b + b - a = 9 + 3
⇒ 2b = 12 ⇒ b = 6
Then a = 9 - 6 = 3
Required number = 10 × 6 + 3 = 63
125.9% ÷ 9.05 x 99.98 = ? - 69.97 × √324.02 ÷ 5.98
85.22 of 499.98% + 299.99 ÷ 30.18 = ?
24.11 × 5.98 + 25.03 × 3.12 – 34.99 + 96.9 × 5.02 =?
1560.182 ÷ √168 + √143 * √224 – 4649.87 ÷ 30.883= ?
570.11 ÷ 18.98 × 5.14 – 123.9 = √?
510.11 ÷ 16.98 × 5.14 – 119.9 = √?
20.02% of (95.96 × 104.01 – 56.02 × 64.04) – ? = 12.02 × 39.96 + 103.03
456.9 + 328.10 - 122.98 = ? + 232.11
15.001% of 799.99 - 3/11% of 1099.99 + 111.002 = ?
(3374 ÷ 125.13)1/3 + (362 ÷ 11) = ?2 – 27.79