Question 14 Marks
There are $m$ boys and $n$ girls in a class. Each boy donated twice as many rupees as is the total number of boys and each girl donated thrice as many rupees as is the total number of girls. The boys collected ₹1152 in all while the girls collected ₹768 in all.
(1) What is the value of $m$ ?
(a) 22$\quad$(b) 24$\quad$(c) 26$\quad$(d) 28
(2) What is the value of $n$ ?
(a) 12$\quad$(b) 14$\quad$(c) 16$\quad$(d) 18
(3) Had there been $(m+3)$ boys and $(n-3)$ girls in the class, the total amount collected would be
(a) ₹ 1925$\quad$(b) ₹ 1940(c) ₹ 1965$\quad$(d) ₹ 2000
(4) Let $p=m^2+n^2$. Then, which of the following is true for $p$ ?
(a) $p$ is a perfect square.(b) $(p+9)$ is a perfect square.(c) $(p+2)$ is a perfect square.(d) $(p-32)$ is a perfect square.
(5) Let $q=m^2-n^2$. Then, the least number to be added to $q$ to make it a perfect square is
(a) 4(b) 6(c) 8(d) 12
(1) What is the value of $m$ ?
(a) 22$\quad$(b) 24$\quad$(c) 26$\quad$(d) 28
(2) What is the value of $n$ ?
(a) 12$\quad$(b) 14$\quad$(c) 16$\quad$(d) 18
(3) Had there been $(m+3)$ boys and $(n-3)$ girls in the class, the total amount collected would be
(a) ₹ 1925$\quad$(b) ₹ 1940(c) ₹ 1965$\quad$(d) ₹ 2000
(4) Let $p=m^2+n^2$. Then, which of the following is true for $p$ ?
(a) $p$ is a perfect square.(b) $(p+9)$ is a perfect square.(c) $(p+2)$ is a perfect square.(d) $(p-32)$ is a perfect square.
(5) Let $q=m^2-n^2$. Then, the least number to be added to $q$ to make it a perfect square is
(a) 4(b) 6(c) 8(d) 12
Answer
View full question & answer→(1) (B) 24
$\begin{array}{l}\text {Amount donated by each boy }=₹(2 m) . \\ \text {Total amount collected by boys }=₹(m \times 2 m)=₹\left(2 m^2\right) . \\ \therefore \quad 2 m^2=1152 \Rightarrow m^2=576 \Rightarrow m=\sqrt{576}=24 .\end{array}$
(2) (C) 16
$\begin{array}{l}\text {Amount donated by each girl }=₹(3 m) . \\ \text {Total amount collected by girls }=₹(n \times 3 n)=₹\left(3 n^2\right) . \\ \therefore \quad 3 n^2=768 \Rightarrow n^2=256 \Rightarrow n=\sqrt{256}=16 .\end{array}$
(3) (C) ₹ 1965
$\begin{array}{l}\text {Number of boys }=27 \text { and number of girls }=13 . \\ \begin{aligned} \text {Total amount collected } & =₹[27 \times(2 \times 27)+13 \times(3 \times 13)] \\ & =₹(2 \times 729+3 \times 169)=₹(1458+507) \\ & =₹ 1965 .\end{aligned}\end{array}$
(4) (B) $(p+9)$ is a perfect square.
$p=m^2+n^2=(24)^2+(16)^2=576+256=832 .$
Now, $(28)^2<832<(29)^2$.
So, next perfect square after $832=(29)^2=841$.
Now, $841-832=9$.
$\therefore(p+9)$ is a perfect square.
(5) (A) 4
$q=m^2-n^2=(24)^2-(16)^2=576-256=320$
Now, $(17)^2<320<(18)^2$.
$\therefore$ least number to be added $=(18)^2-320=324-320=4$.
$\begin{array}{l}\text {Amount donated by each boy }=₹(2 m) . \\ \text {Total amount collected by boys }=₹(m \times 2 m)=₹\left(2 m^2\right) . \\ \therefore \quad 2 m^2=1152 \Rightarrow m^2=576 \Rightarrow m=\sqrt{576}=24 .\end{array}$
(2) (C) 16
$\begin{array}{l}\text {Amount donated by each girl }=₹(3 m) . \\ \text {Total amount collected by girls }=₹(n \times 3 n)=₹\left(3 n^2\right) . \\ \therefore \quad 3 n^2=768 \Rightarrow n^2=256 \Rightarrow n=\sqrt{256}=16 .\end{array}$
(3) (C) ₹ 1965
$\begin{array}{l}\text {Number of boys }=27 \text { and number of girls }=13 . \\ \begin{aligned} \text {Total amount collected } & =₹[27 \times(2 \times 27)+13 \times(3 \times 13)] \\ & =₹(2 \times 729+3 \times 169)=₹(1458+507) \\ & =₹ 1965 .\end{aligned}\end{array}$
(4) (B) $(p+9)$ is a perfect square.
$p=m^2+n^2=(24)^2+(16)^2=576+256=832 .$
Now, $(28)^2<832<(29)^2$.
So, next perfect square after $832=(29)^2=841$.
Now, $841-832=9$.
$\therefore(p+9)$ is a perfect square.
(5) (A) 4
$q=m^2-n^2=(24)^2-(16)^2=576-256=320$
Now, $(17)^2<320<(18)^2$.
$\therefore$ least number to be added $=(18)^2-320=324-320=4$.