8. Six bells commence tolling together and toll at intervals of 2,4,6,8,10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
a. 4
b. 10
c. 15
d. 16
Answer: D
Explanation:
$\qquad \begin{array}{|l}
\llap{2~~~~} 2,4,6,8,10, 12 \\[4px] \hline
\llap{2~~~~} 1, 2, 3, 4, \phantom{0}5, \phantom{0}6 \\[4px] \hline
\llap{3~~~~} 1, 1, 3, 2, \phantom{0}5,\phantom{0}3 \\[4px] \hline
\llap{~~~~} 1, 1, 1, 2, \phantom{0}5,\phantom{0}1 \\[4px]
\end{array}$
LCM of 2,4,6, 8,10 and 12 = $ = {2^3} \times 3 \times 5$ $= 120$.
So, the bells will toll simultaneously after 120 seconds, i.e.2 minutes.
In 30 minutes, they $\left( {\displaystyle\frac{{30}}{2} + 1} \right)$ toll times ie.16 times.
9. LCM of 87 and 145 is :
a. 870
b. 1305
c. 435
d. 1740
10. The traffic lights at three different road crossing change after every 48 sec; 72 sec; and 108 sec., respectively. If they all change simultaneously at 8:20:00 hrs, then they will again change simultaneously at
a. 8:27:12 Hrs
b. 8:27:24 Hrs
c. 8:27:36 Hrs
d. 8:27:48 Hrs
11. The greatest number by which if 1657 and 2037 are divided the remainders will be 6 and 5 respectively is
a. 127
b. 235
c. 260
d. 305
Answer: A
Explanation:
The needed number is HCF of $(1657-6)$ and $(2037-5)$ $= \text{HCF} (1651 ,2032)$
$\quad\require{enclose}
\begin{array}{rll}
1651 ){2032}(1{\phantom{00000000}} \\[2px]
\underline{1651}{\phantom{0000000000}} \\[1px]
381) 1651(4 {\phantom{0000}}\\[1px]
\underline {1524}{\phantom{000000}} \\[1px]
\boxed{127}\phantom{})381(3{\phantom{}} \\[1px]
\underline{381}\phantom{00} \\[1px]
0\phantom{00}
\end{array}$
12. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divisible by 3, then the first number is
a. 264
b. 132
c. Both a and b
d. 33
Answer: C
Explanation:
Let the numbers are $Ah$, $Bh$ respectively. Here $h$ is $HCF$ of two numbers. (obviously $A, B$ are coprimes i.e., $HCF (A, B) = 1$)
Given that $HCF = h = 44$ and $LCM = ABh = 264$
Dividing LCM by HCF we get $AB = 6$.
AB can be written as $1 \times 6, 2 \times 3, 3 \times 2, 6 \times 1$.
But given that the first number is divisible by 3.
So only two options possible for A.
$3 \times 44 = 132, 6 \times 44 = 264$.
So option C is correct.
13. What least number must be subtracted from 1294 so that the remainder when divided 9, 11, 13 will leave in each case the same remainder 6 ?
a. 0
b. 1
c. 2
d. 3
Answer: B
Explanation:
LCM of 9,11,13 is 1287. Dividing 1294 with 1287, the remainder will be 7, to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.
14. The least number which is divisible by 12, 15, 20 and is a perfect square, is
a. 400
b. 900
c. 1600
d. 3600
Answer: B
Explanation:
$\qquad \begin{array}{|l}
\llap{5~~~~} 12, 15, 20 \\[4px] \hline
\llap{4~~~~} 12, \phantom{0}3, \phantom{0}4 \\[4px] \hline
\llap{3~~~~} \phantom{0}3, \phantom{0}3, \phantom{0}1 \\[4px] \hline
\llap{~~~~} \phantom{0}1, \phantom{0}1, \phantom{0}1 \\[4px]
\end{array}$
LCM $= 5 × 3 × 2 × 2 = {{\text{2}}^2} \times 3 \times 5 = 60$
To make this number as a perfect square, we have to multiply this number by $5 × 3$
The number is $60 × 15= 900$
