Aptitude Problems On NumberPage 3
17. | | In a two digit number, If it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is? | | (a)21 | | (b)22 | | (c)23 | | (d)24 |
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Answer is: DLet, the ten's digit be X. Then, unit's digit = X+2
Number = 10X + (X+2) = 11X + 2
Sum of digits X + X + 2 = 2X + 2
∴ (11X + 2)(2X + 2) = 144
∴ 22X² + 26X – 140 = 0
⇒ 11X² + 13X – 70 = 0
⇒ (X - 2)(11X +35) = 0
⇒ X = 2
Hence, required number = 11X + 2 = 24
18. | | In a two digit number, the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged, difference between the newly formed number and the original number is less than the original number by 1. What is the original number? | | (a)37 | | (b)38 | | (c)39 | | (d)40 |
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Answer is: ALet, the ten's digit be X. Then, unit's digit = 2X+1
[10X + (2X + 1)] – [{10(2X + 1) + X} – {10X + (2X +1)}] = 1
⇒ (12X + 1) – (9X +9) = 1
⇒ 3X = 9
⇒ X = 3
So, ten's digit = 3 and unit's digit = 7
Hence, original number = 37
19. | | A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is? | | (a)26 | | (b)27 | | (c)28 | | (d)29 |
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Answer is: BLet the ten's digit be X and unit's digit be Y.
Then, 10X + Y = 3(X + Y)
⇒ 7X – 2Y = 0 ….......(1)
10X + Y + 45 = 10Y + X
⇒ Y – X = 5 …………(2)
Solving (1) and (2), We get : X = 2 and Y = 7.
∴ Required number = 27.
20. | | Of the three numbers, the sum of the first two is 45, the sum of the second and third is 55 and the sum of the third and thrice the first is 90. The third number is? | | (a)30 | | (b)31 | | (c)32 | | (d)33 |
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Answer is: ALet, the number be X, Y and Z.
Then, X + Y = 45, Y + Z = 55 and 3X + Z = 90
⇒ Y = 45 – X, Z = 55 – Y = 55 – (45 - X) = 10 + X.
∴ 3X + 10 + X = 90 or X = 20.
∴ Y = (45 – 20) = 25 and Z = (10 + 20) = 30
∴ Third number = 30.
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