AD Practice SET – 5
Age & Digits: Level 5
1.
Find the number of two-digit numbers \( x \) such that the product of the digits of \( x \) is half of \( x \).
Solution (C):
Let \( x = 10t + u \).
Condition: \( t \times u = \frac{1}{2}(10t + u) \).
\( 2tu = 10t + u \).
\( 2tu – u = 10t \implies u(2t-1) = 10t \).
\( u = \frac{10t}{2t-1} \).
Since \( u \) is a single digit (0-9), we test values of \( t \) (1-9):
\( t=1 \implies u = 10/1 = 10 \) (Not a digit).
\( t=2 \implies u = 20/3 \) (Not integer).
\( t=3 \implies u = 30/5 = 6 \). (Number 36. Check: \(3 \times 6 = 18\). \(36/2 = 18\). Valid).
\( t=4 \implies u = 40/7 \) (No).
\( t=5 \implies u = 50/9 \) (No).
Higher \( t \) yields smaller \( u \) but not integers.
Only 1 solution: 36.
2.
In a family, the average age of the father and mother is 35 years. The average age of the father, mother, and their only son is 27 years. What is the age of the son?
Solution (D):
Sum(F + M) = \( 35 \times 2 = 70 \).
Sum(F + M + S) = \( 27 \times 3 = 81 \).
Age of Son = \( 81 – 70 = 11 \).
3.
When 1 is added to the product of two prime numbers, the result is a perfect square. How many such pairs of primes less than 20 exist?
Solution (E):
We want \( p_1 p_2 + 1 = k^2 \implies p_1 p_2 = k^2 – 1 = (k-1)(k+1) \).
Since \(p_1, p_2\) are prime, one of the factors must be 2 (because primes are integers > 1, and the only consecutive even/odd breakdown requires 2. Actually, one factor could be 1, but primes > 1).
Actually, \(k-1\) and \(k+1\) differ by 2.
Case A: \(k-1 = 2 \implies k=3\). Then \(p_1 p_2 = (2)(4) = 8\) (Not product of two primes).
Wait, \(p_1 p_2\) must match the factors.
If \(p_1 = k-1\) and \(p_2 = k+1\): Both must be prime. (Twin primes separated by 2).
Pairs < 20:
(3, 5): \(3 \times 5 + 1 = 16 = 4^2\). (Yes).
(5, 7): \(5 \times 7 + 1 = 36 = 6^2\). (Yes).
(11, 13): \(11 \times 13 + 1 = 144 = 12^2\). (Yes).
(17, 19): \(17 \times 19 + 1 = 324 = 18^2\). (Yes).
Is 2 a part? \(2 \times p + 1 = k^2\).
If p=3, \(7\) (No). p=5, \(11\). p=7, \(15\). p=11, \(23\).
So only twin primes work.
Pairs < 20 means primes themselves are < 20.
(3,5), (5,7), (11,13), (17,19).
That's 4 pairs.
Wait, option E is 3. D is 4.
Let me recheck "less than 20".
Primes: 2, 3, 5, 7, 11, 13, 17, 19.
Pairs: (3,5), (5,7), (11,13), (17,19).
All valid. So 4 pairs.
I will set Answer to **D (4)**.
(Correction: Code below has E, I will fix to D).
3.
When 1 is added to the product of two prime numbers, the result is a perfect square. How many such pairs of primes less than 20 exist?
Solution (D):
Equation: \(xy + 1 = k^2 \implies xy = k^2 – 1 = (k-1)(k+1)\).
This implies x and y are twin primes (differ by 2).
Pairs < 20: (3,5), (5,7), (11,13), (17,19).
Total 4 pairs.
4.
My brother is 3 years older than me. My father was 28 years of age when my sister was born, while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, what was the age of my father and mother respectively when my brother was born?
Solution (B):
Sister born: Father = 28.
Brother born: Sister was 4. So Father = \( 28 + 4 = 32 \).
Me born: Mother = 26.
Brother is 3 years older than Me.
When Brother was born, “Me” was -3 years old (3 years before birth).
So Mother was \( 26 – 3 = 23 \).
Father: 32, Mother: 23.
5.
A person is asked his age. He says, “Take 3 times my age 3 years hence, subtract 3 times my age 3 years ago, and you will know how old I am.” What is his age?
Solution (A):
Let age = \( x \).
\( 3(x+3) – 3(x-3) = x \).
\( 3x + 9 – 3x + 9 = x \).
\( 18 = x \).
Score: 0 / 0