You visited us 1 times! Enjoying our articles? Unlock Full Access!

Formation of Algebraic Expressions

Q. Consider t...

Question

Q. Consider the following multiplication problem:

(PQ) x 3 = RQQ, where P, Q and R are different digits and R ≠ 0.

What is the value of (P + R) ÷ Q?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Cannot be determined

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 5
Here $Q \times 3 = a$ no. which has ‘Q’ in its unit’s place, and this happens only for $P = 5$ (as 5 x 3 = 15). Also, $P \times 3 + 1$ (1 is carryover from 15) = R5. This will happen only when $P = 8$ (as 8 x 3 + 1 = 25). Hence, $P = 8, R = 2$ and $Q = 5.$ Now $(8 + 2) \div 5 = 10 \div 2 = 5.$

Suggest Corrections

Similar questions

{log}_{p} q + {log}_{q} r + {log}_{r} p

p, q

r

{({log}_{p} q)}_{p}^{3} + {({log}_{q} r)}_{q}^{3} + {({log}_{r} p)}_{r}^{3}

\frac{(x + q) (x + r)}{(q - p) (r - p)} + \frac{(x + r) (x + p)}{(r - q) (p - q)} + \frac{(x + p) (x + q)}{(p - r) (q - r)} - 1 = 0

p, q, r

P =

3

Q =

4

R =

6

1

P \cup Q = R

2.

P \subset R

3.

R \subset (P \cup Q)

p, q, r

(q - r) x + (r - p) y + (p - q) = 0

(q^{3} - r^{3}) x + (r^{3} - p^{3}) y + (p^{3} - q^{3}) = 0

P : [(⇁ p \lor q) \land (r \to s) \land (p \lor r)] \to (⇁ s \to q)

Q : [(⇁ p \land q) \land [q \to (p \to r)]] \to⇁ r

R : [[(q \land r) \to p] \land (⇁ q \lor p)] \to r

S : [p \land (p \to r) \land (q \lor ⇁ r)] \to q

Join BYJU'S Learning Program