MCQ
The sum of three consecutive multiples of $7$ is $357.$ Find the smallest multiple.
- ✓$112$
- B$126$
- C$119$
- D$116$
✓
Answer
Correct option: A.
$112$
Let the three consecutive multiplies of $7$ be $7x, (7x + 7), (7x + 14)$ where $x$ is a natural number.
According to question,
$7x + (7x + 7) + (7x + 14) = 357$
$21x + 21 = 357$
$21(x + 1) = 357$
$\frac{21(\text{x}+1)}{21}=\frac{357}{21}$
$x + 1 = 17$
$x = 17 - 1$
$x = 16$
Hence, the smallest multiple of $7$ is $7 × 16$ i.e., $112.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$(0.8)^3=?$
→The coefficient of $x^2y$ in $7pqrx^2$ is:
→The product of $5x$ and $2y$ is
→A bag is available for Rs. $90$. The shopkeeper allows $10 \%$ discount on the marked price. What is the marked price of the bag?
→One $(1)$ is:
→If $x=(50)^{1-6}+(50)^{-2}$, then the value of $x^{-3}$ is
→If $\text{PQ}$ and $\text{RS}$ are two perpendicular diameters of a circle, then $\text{PQRS}$ is a:
→What must be added to $-\frac{5}{16}$ to get $\frac{5}{8}.$
→Observe the following runs-over graph and answer the related questions:
In which over are the maximum runs scored?
In which over are the maximum runs scored?
If $\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$, then $n=?$
→