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1. If one-third of one-fourth of a number is 15, then three-tenth of that number is:

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Solution:
LetthenumberbexThen,13of14ofx=15⇒x=15×12=180So,requirednumber=310×180=54
2. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

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Solution:
Let the three integers be x, x + 2 and x + 4
Then, 3x = 2(x + 4) + 3    ⇔    x = 11
∴ Third integer = x + 4 = 15
3. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

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Solution:
Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 36
⇒ 9(x - y) = 36
⇒ x - y = 4
4. The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

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Solution:
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 × 2x + x) - (10x + 2x) = 36
⇒ 9x = 36
⇒ x = 4
∴ Required difference
= (2x + x) - (2x - x)
= 2x
= 8
5. A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

Discuss
Solution:
Let the ten's and unit digit be x and 8x respectively
Then,(10x+8x)+18=10×8x+x⇒10x2+8+18x=80+x2⇒9x2+18x−72=0⇒x2+2x−8=0⇒(x+4)(x−2)=0⇒x=2
∴ first digit will be 2 and second digit will be 4.
i.e digit is 24.
6. The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

Discuss
Solution:
Let the ten's digit be x and unit's digit be y
Then, x + y = 15 and x - y = 3   or   y - x = 3
Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6
Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69
Hence, the number cannot be determined.
7. The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

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Solution:
Let the numbers be a, b and c
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400
⇒ (a + b + c) = 400  = 20
8. A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

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Solution:
Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y
Number obtained by interchanging the digits = 10y + x
∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11
9. In a two-digit, 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:

Discuss
Solution:
Let 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
⇒ 11x2 + (35 - 22)x - 70 = 0
⇒ 11x2 + 35x - 22x - 70 = 0
⇒ (x - 2)(11x + 35) = 0
⇒ x = 2
Hence, required number = 11x + 2 = 24
10. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

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Solution:
LetthenumberbexThen,x+17=60x⇒x2+17x−60=0⇒(x+20)(x−3)=0⇒x=3

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