16. Three piece of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?
a. 7 m
b. 14 m
c. 42 m
d. 63 m
Answer: A
Explanation:
The maximum possible length = (HCF of 42, 49, 63)
${\begin{array}{*{20}{cl}}
{42}&{ = 2^1 \times 3^1 \times 7^1} \\[2px]
{49}&{ = {7^2}} \\[2px]
{63}&{ = {3^2 \times 7^1}}
\end{array}}$
HCF of $42, 49, 63 = 7$
($\because $ HCF of the above numbers is the $\text{minimum power}$ $\text{of common prime number}$)
17. The greatest number which can divide 1354, 1866, 2762 leaving the same remainder 10 in each case is :
a. 64
b. 124
c. 156
d. 260
18. A teacher when distributed certain number of chocolates to 4 children, 5 children, 7 children, left with 1 chocolate. Find the least number of chocolates the teacher brought to the class
a. 101
b. 111
c. 121
d. 141
Answer: D
Explanation:
$N = K (LCM (4, 5, 7) + 1$ $= 140K + 1$ where $K$ = natural number.
When we substitute $K = 1$, we get the least number satisfies the condition.
So minimum chocolates $= 141$
19. When certain number of marbles are divided into groups of 4, one marble remained. When the same number of marbles are divided into groups of 7 and 12 then 4, 9 marbles remained. If the total marbles are less than 10,000 then find the maximum possible number of marbles.
a. 9116
b. 9987
c. 9991
d. 9993
Answer: D
Explanation:
In this case the difference between the remainders and divisors is constant. i.e., 3.
so $N = K (LCM (4, 7, 12) - 3$ $= 84K - 3$ where $K$ = natural number.
But we know that $84K - 3$ < 10,000
$\Rightarrow 84 \times 119 - 3 < 10,000$
$ \Rightarrow $ $9996 - 3 = 9993$
20. Find the greatest number, which will divide 260, 281 and 303, leaving 7, 5 and 4 as remainders respectively.
a. 21
b. 22
c. 23
d. 27
Answer: C
Explanation:
We have to find the HCF of (260 - 7, 281 - 5, 303 - 4) = HCF (253, 276, 299)
HCF(253, 276) =
$\quad\require{enclose}
\begin{array}{rll}
253 ){276}(1{\phantom{00000}} \\[2px]
\underline{253}{\phantom{0000000}} \\[1px]
\boxed{23})253(11{\phantom{0}}\\[1px]
\underline {23}{\phantom{00000}} \\[1px]
23\phantom{}{\phantom{0000}} \\[1px]
\underline {23}{\phantom{0000}} \\[1px]
0\phantom{}{\phantom{0000}} \\[1px]
\end{array}$
