20. By what number $a.pq\,pq\,pq...$ should be multiplied so that it becomes an integer?
a. 90
b. 99
c. 100
d. 1000000
Answer: B
Explanation:
$\begin{align}
\begin{aligned}
a.pq\,pq\,pq\, . . . &= a.\overline {pq} \\[6px]
&= a + \dfrac{{pq}}{{99}} \\[6px]
&= \dfrac{{99a+pq}}{{99}} \\[6px]
\end{aligned}
\end{align}$
Therefore, if we multiply the above fraction by $99$ or any multiple of $99$ it will become an integer.
21. By what number $a.b\,pq\,pq\,pq\,...$ should be multiplied so that it becomes an integer?
a. 900
b. 990
c. 1980
d. Both b and c
Answer: D
Explanation:
$\begin{align}
\begin{aligned}
a.b\,pq\,pq\,pq\, . . . & = a.b\,\overline {pq} \\[6px]
&= a + \dfrac{{bpq-b}}{{990}} \\[6px]
&= \dfrac{{990a+bpq-b}}{{990}} \\[6px]
\end{aligned}
\end{align}$
Therefore, if we multiply the above fraction by $990$ or any multiple of $990$ it will become an integer.
22. Sum of the numbers from 1 to 20 is
a. 210
b. 110
c. 220
d. 105
Answer: A
Explanation:
Sum of first n natural numbers = 1 + 2 + 3 + ..... n = $\displaystyle\frac{{n(n + 1)}}{2}$
Substitute n = 20.
So \({S_{20}} = \dfrac{{20 \times 21}}{2} = 210\)
23. Sum of even numbers between 15 and 25 is
a. 70
b. 80
c. 130
d. 100
Answer: D
Explanation:
16 + 18 + .......24.
Taking 2 common we get = 2 ( 8 + 9 + 10 +.....+ 12)
Sum of n natural numbers upto 12
8 + 9 + 10 +.....+ 12 = (1 + 2 + 3 + ......+ 12) - ( 1 + 2 + 3 + ....+ 7)
By applying the formula for the first n natural numbers sum $\displaystyle\frac{{n(n + 1)}}{2}$ we get, $\displaystyle\frac{{12(12 + 1)}}{2} - \displaystyle\frac{{7(7 + 1)}}{2} = 50$
So 16 + 18 + .......24 = 2 × 50 = 100
Alternative Method:
16 + 18 + .......24
This is an Arithmetic progression with $a = 16, d = 2, l = 24$
Total terms in the sequence is given by = \(n = \dfrac{{l - a}}{d} + 1\)
So total terms = \(n = \dfrac{{24 - 16}}{2} + 1 = 5\)
Sum of the terms when first term and last term in known = \(\dfrac{n}{2}\left( {a + l} \right)\) = \(\dfrac{5}{2}\left( {16 + 24} \right) = 100\)
24. How many numbers between 1000 and 5000 are exactly divisible by 225?
a. 16
b. 18
c. 19
d. 12
Answer: B
Explanation:
First multiple of 225 after 1000 is 1125 (225 × 5 ) and last multiple of 225 before 5000 is 4950 (225 × 22)
Total number are $\displaystyle\frac{{l - a}}{d} + 1 = \displaystyle\frac{{4950 - 1125}}{{225}} + 1 = 18$
