9th Class ICSE Mathematics Question Paper

SECTION A

Attempt All the Question (4×10=40)

$1 a) If x=1–\sqrt{2}, Find the value of (x–\frac{1}{x}{)}^3$ $b) A borrowed Rs 2500 from B at 12\% per annum compound interest. After 2 Years A gave Rs 2936$
$and a watch to B to clear the Account. Find the cost of Watch$
$C) Find the amount and the compoud interest on Rs 12000 in 3 years$
$at 5\% interest being compounded yearly.$

$2 a) If a^2–5a–1=0 and a\ne 0 find$
$\left(i\right) a–\frac{1}{a} \left(ii\right) a+\frac{1}{a} \left(iii\right) a^2–\frac{1}{a^2}$
$b) Factorise : a^2 + 10a+24$ $c) If AP bisects angle BAC and M is any point on AP .Prove that perpendicular drawn from$
$M to AB and AC are equal$

$3 \mathrm{a}) \mathrm{Find} \mathrm{the} \mathrm{actual} \mathrm{lower} \mathrm{and} \mathrm{upper} \mathrm{class} \mathrm{limits} \mathrm{and} \mathrm{also} \mathrm{the} \mathrm{class} \mathrm{marks} \mathrm{of} \mathrm{the} \mathrm{classes}$
$1.1–2.0 \mathrm{;} 2.1–3.0 \mathrm{;} 3.1–4.0$
$\mathrm{b}) \mathrm{Find} \mathrm{the} \mathrm{mean} \mathrm{of} \mathrm{all} \mathrm{factors} \mathrm{of} 10$
$\mathrm{c}) \mathrm{Simplify} \frac{\sqrt{2}}{\sqrt{6}–\sqrt{2}}–\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$4 \mathrm{a}) \mathrm{The} \mathrm{difference} \mathrm{between} \mathrm{the} \mathrm{simple} \mathrm{interest} \mathrm{and} \mathrm{compound} \mathrm{interest} \mathrm{on} \mathrm{a} \mathrm{certain} \mathrm{sum} \mathrm{is} \mathrm{Rs} 54.40$
$\mathrm{for} 2 \mathrm{years} \mathrm{at} 8\% \mathrm{per} \mathrm{annum}.\mathrm{Find} \mathrm{the} \mathrm{sum}$
$\mathrm{b}) \mathrm{Calculate} \mathrm{the} \mathrm{amount} \mathrm{and} \mathrm{the} \mathrm{compound} \mathrm{interest} \mathrm{on} \mathrm{Rs} 8000 \mathrm{in} 2.5 \mathrm{years} \mathrm{at} 15\% \mathrm{per} \mathrm{annum}$
$\mathrm{c}) \mathrm{If} ( \mathrm{a}+\frac{1}{\mathrm{a}}{)}^2=3 \mathrm{and} \mathrm{a}\ne 0, \mathrm{then} \mathrm{show} \mathrm{that} {\mathrm{a}}^3+\frac{1}{{\mathrm{a}}^3}=0$

SECTION B

Do any Four Question (4×10=40)

$5 a) Factorise : 50a^3–2a$
$\mathrm{b}) \mathrm{The} \mathrm{Perpendicular} \mathrm{bi}sec\mathrm{tor} \mathrm{of} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{a} △\mathrm{ABC} \mathrm{meet} \mathrm{at} \mathrm{I}.$
$\mathrm{Prove} \mathrm{that} \mathrm{IA}=\mathrm{IB}=\mathrm{IC}$
$\mathrm{c}) \mathrm{Construct} \mathrm{a} \mathrm{frequency} \mathrm{polygon} \mathrm{for} \mathrm{the} \mathrm{following} \mathrm{distributions}$

Class – Internals	0 – 4	4 – 8	8 – 12	12 – 16	16 – 20	20 – 24
Frequency	4	7	10	15	11	6

$6 \mathrm{a}) \mathrm{If} \mathrm{the} \mathrm{mean} \mathrm{of} \mathrm{observations} \mathrm{x},\mathrm{x}+2,\mathrm{x}+2,\mathrm{x}+4,\mathrm{x}+6 \mathrm{and} \mathrm{x}+8 \mathrm{is} \mathrm{mean}$
$\mathrm{find} \mathit{\left(}i\mathit{\right)}\mathit{ }the\mathit{ }value\mathit{ }of\mathit{ }x$
$\left(\mathrm{ii}\right)\mathrm{the} \mathrm{mean} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{three} \mathrm{observations}$
$\mathrm{b}) \frac{5–\sqrt{7}}{5+\sqrt{7}}–\frac{5+\sqrt{7}}{5–\sqrt{7}} is equal to$
$\left(\mathrm{i}\right) 10\sqrt{7} \left(\mathrm{ii}\right) \frac{1\sqrt{7}}{9}$
$\left(\mathrm{iii}\right) \frac{10\sqrt{7}}{9} \left(\mathrm{iv}\right) \frac{–10\sqrt{7}}{9}$
$\mathrm{c}) \mathrm{Calculate} \mathrm{the} \mathrm{amount} \mathrm{and} \mathrm{compound} \mathrm{interest} \mathrm{on}: \mathrm{Rs} 4600 \mathrm{in} 2 \mathrm{years}$
$\mathrm{when} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{interests} \mathrm{of} \mathrm{successive} \mathrm{years} \mathrm{are} 10\% \mathrm{\&} 12\% \mathrm{respectively}$


$7 \mathrm{a}) \mathrm{Find} \mathrm{the} \mathrm{difference} \mathrm{between} \mathrm{the} \mathrm{compound} \mathrm{interest} \mathrm{compounded} \mathrm{yearly} \mathrm{and}$
$\mathrm{half} \mathrm{yearly} \mathrm{on} \mathrm{Rs} 10000 \mathrm{for} 18 \mathrm{months} \mathrm{at} 10\% \mathrm{per} \mathrm{annum}$
$\mathrm{b}) \mathrm{If} \mathrm{x}+\mathrm{y}–\mathrm{z}=4 \mathrm{and} {\mathrm{x}}^2+{\mathrm{y}}^2+z^2=30 then find the value of$
$\mathrm{xy}–\mathrm{yz}–\mathrm{zx}$
$\mathrm{c}) \mathrm{Factorize} : {\mathrm{a}}^6–7{\mathrm{a}}^3–8$

$8 \mathrm{a}) \mathrm{A} \mathrm{line} \mathrm{segement} \mathrm{AB} \mathrm{is} \mathrm{bi}\mathrm{sected} \mathrm{at} \mathrm{point} \mathrm{P} \mathrm{and} \mathrm{through} \mathrm{point} \mathrm{P} \mathrm{another}$
$\mathrm{line} \mathrm{segement} \mathrm{PQ} \mathrm{which} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{AB} \mathrm{is} \mathrm{drawn} \mathrm{show} \mathrm{that} \mathrm{QA}=\mathrm{QB}$
$\mathrm{b}) \mathrm{The} \mathrm{median} \mathrm{of} \mathrm{Observations} 10,11,13,17,\mathrm{x}+5,20,22,24 \mathrm{and} 53$
$(\mathrm{arranged} \mathrm{in} \mathrm{ascending} \mathrm{order}) \mathrm{is} 18. \mathrm{Find} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{x}.$
$\mathrm{c}) \mathrm{The} \mathrm{marks} \mathrm{of} 24 \mathrm{candidates} \mathrm{in} \mathrm{the} \mathrm{subject} \mathrm{Mathematics} \mathrm{are} \mathrm{given} :$

45	48	15	23	30	35	40	11
29	0	03	12	48	50	18	30
15	30	11	42	28	02	03	`44

The maximum marks are 50. Make a frequency distribution taking class intervals 0-10,10-20,—–

$9 \mathrm{a}) \mathrm{x}=2\sqrt{3}+2\sqrt{2} find:$
$\left(i\right) \frac{1}{x} \left(ii\right) x+\frac{1}{x} \left(iii\right) (x+\frac{1}{x}{)}^2$
$\mathrm{b}) \mathrm{How} \mathrm{much} \mathrm{will} \mathrm{Rs} 50000 \mathrm{amount} \mathrm{to} \mathrm{in} 3 \mathrm{years} \mathrm{compounded} \mathrm{yearly}, \mathrm{if} \mathrm{the} \mathrm{rate} \mathrm{for}$
$\mathrm{the} \mathrm{successive} \mathrm{years} \mathrm{are} 6\%,8\% \mathrm{\&} 10\% \mathrm{respectively}.$
$\mathrm{c}) \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{an} \mathrm{article} \mathrm{decreased} \mathrm{for} \mathrm{two} \mathrm{years} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{of} 10\% \mathrm{per} \mathrm{year} \mathrm{and} \mathrm{then} \mathrm{in} \mathrm{the}$
$\mathrm{third} \mathrm{year} \mathrm{it} \mathrm{increased} \mathrm{by} 10\%. \mathrm{Find} \mathrm{the} \mathrm{original} \mathrm{value} \mathrm{of} \mathrm{article}, \mathrm{if} \mathrm{its} \mathrm{value} \mathrm{at} \mathrm{the} \mathrm{end} \mathrm{of} 3 \mathrm{years}$
$\mathrm{is} \mathrm{Rs} \mathrm{40,095.}$

$10 \mathrm{a}) \mathrm{If} \mathrm{a}+\frac{1}{\mathrm{a}}=6 \mathrm{and} \mathrm{a}\ne 0 \mathrm{find}:$
$\left(\mathrm{i}\right) \mathrm{a}–\frac{1}{\mathrm{a}} \left(\mathrm{ii}\right) {\mathrm{a}}^2–\frac{1}{{\mathrm{a}}^2}$
$\mathrm{b}) \mathrm{Factorise}: (3\mathrm{x}–2\mathrm{y}{)}^2 +3(3\mathrm{x}–2\mathrm{y})–10$
$\mathrm{c}) \mathrm{The} \mathrm{Figure} \mathrm{show} \mathrm{a} \mathrm{circle} \mathrm{with} \mathrm{centre} \mathrm{O}. \mathrm{If} \mathrm{OP} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{AB}.$
$\mathrm{Prove} \mathrm{that} \mathrm{AP}=\mathrm{BP}$

$11 \mathrm{a}) \mathrm{the} \mathrm{mean} \mathrm{of} 5 \mathrm{numbers} \mathrm{is} 18.\mathrm{if} \mathrm{one} \mathrm{number} \mathrm{is} \mathrm{exceluded},\mathrm{the} \mathrm{mean} \mathrm{of} \mathrm{remaining}$
$\mathrm{numbers} \mathrm{becomes} 16.\mathrm{find} \mathrm{the} \mathrm{exceluded} \mathrm{number}$
$\mathrm{b}) \mathrm{Factorise}: {\mathrm{a}}^2 –{\mathrm{b}}^2–(\mathrm{a}+\mathrm{b}{)}^2$
$\mathrm{c})\mathrm{if} \mathrm{x}\ne 0 \mathrm{and} \mathrm{x}–\frac{1}{\mathrm{x}}=2 \mathrm{then} \mathrm{show} \mathrm{that}:$
${\mathrm{x}}^2 +\frac{1}{{\mathrm{x}}^2}={\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3}={\mathrm{x}}^4 +\frac{1}{{\mathrm{x}}^4}$