Daily Math Problems — June 19, 2026: Logaritmi, Determinants, Probability

dailymath

Problems & worked solutions

RO M1

Problem 1 — Logarithms

Calculați $\log_{5} 5^{12} - \log_{7} 1$ .

Show answer & worked solution

A. $0$
B. $5$
C. $11$
D. $12$ ✓ correct

Avem $\log_{5} 5^{12} = 12$ și $\log_{7} 1 = 0$ . Deci expresia este $12 - 0 = 12$ .

RO M1

Problem 2 — Determinants

A square matrix $A$ is invertible iff:

Show answer & worked solution

A. $\det (A) = 0$
B. $A is symmetric$
C. $\det (A) \neq 0$ ✓ correct
D. $all entries are non-zero$

Invertibility ⇔ non-zero determinant.

RO M1

Problem 3 — Probability

Two boxes — Box A has $3$ red and $1$ blue; Box B has $1$ red and $3$ blue. A box is chosen at random and a ball is drawn. Given the ball is red, the probability it came from Box A is:

Show answer & worked solution

A. $\frac{1}{4}$
B. $\frac{1}{2}$
C. $\frac{3}{4}$ ✓ correct
D. $\frac{3}{8}$

$P (A) = P (B) = \frac{1}{2}$ , $P (R ∣ A) = \frac{3}{4}$ , $P (R ∣ B) = \frac{1}{4}$ .

$P (A ∣ R) = \frac{\frac{3}{4} \cdot \frac{1}{2}}{\frac{3}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2}} = \frac{3 / 8}{4 / 8} = \frac{3}{4}$ .

Problems & worked solutions

RO M1

Problem 1 — Logarithms

Calculați $\log_{5} 5^{12} - \log_{7} 1$ .

Show answer & worked solution

A. $0$
B. $5$
C. $11$
D. $12$ ✓ correct

Avem $\log_{5} 5^{12} = 12$ și $\log_{7} 1 = 0$ . Deci expresia este $12 - 0 = 12$ .

RO M1

Problem 2 — Determinants

A square matrix $A$ is invertible iff:

Show answer & worked solution

A. $\det (A) = 0$
B. $A is symmetric$
C. $\det (A) \neq 0$ ✓ correct
D. $all entries are non-zero$

Invertibility ⇔ non-zero determinant.

RO M1

Problem 3 — Probability

Two boxes — Box A has $3$ red and $1$ blue; Box B has $1$ red and $3$ blue. A box is chosen at random and a ball is drawn. Given the ball is red, the probability it came from Box A is:

Show answer & worked solution

A. $\frac{1}{4}$
B. $\frac{1}{2}$
C. $\frac{3}{4}$ ✓ correct
D. $\frac{3}{8}$

$P (A) = P (B) = \frac{1}{2}$ , $P (R ∣ A) = \frac{3}{4}$ , $P (R ∣ B) = \frac{1}{4}$ .

$P (A ∣ R) = \frac{\frac{3}{4} \cdot \frac{1}{2}}{\frac{3}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2}} = \frac{3 / 8}{4 / 8} = \frac{3}{4}$ .