Q&A: practice questions with answers

Concept: A series (1, 2, 3… or Jan, Feb… or a custom list) is created by typing the first value(s), selecting them and dragging the Fill Handle (small square at the cell's bottom-right corner).

1. Custom lists are defined in File → Options → Advanced → Edit Custom Lists, then filled by dragging the Fill Handle.
2. Data Validation restricts entries, Conditional Formatting colours cells, Sort & Filter arranges data – none generate a series.

Yaad rakhein: Fill Handle = AutoFill = series.

Answer: (D) Using the Fill Handle and dragging

Concept: In Outlook's message window, Style Sets are listed under Change Styles → Style Set in the Styles group, not under the Styles Dialog Box Launcher.

1. Statements A (Change Styles → Colors/Fonts), B (Style Inspector at the bottom of the Styles pane) and C (live preview on hovering) are correct Microsoft procedures.
2. Statement D is false: the Dialog Box Launcher opens the Styles pane for individual styles; the list of style sets is under Change Styles.

Yaad rakhein: Style Set → Change Styles button; Styles pane → individual styles.

Answer: (D) After you select a Style, you can see a list of the style sets available within the styles…

Concept: A file infector virus attaches its code to executable files (.exe, .com) and runs – and spreads – whenever the infected program is executed.

1. Macro virus: hides in Office document macros. Boot sector virus: infects the disk's boot record and loads at start-up.
2. Polymorphic virus: changes its code each time to evade antivirus – a disguise technique, not a target type.

Yaad rakhein: 'File infector' – the name says what it infects.

Answer: (D) File Infector Virus

Concept: A LAN covers a small area (office, campus) and offers high speed (100 Mbps–10 Gbps) – that is an advantage.

1. Disadvantages: high initial cost of cabling, switches and servers; limited geographical size; security risk if the central server is compromised.
2. High data transmission rate is a benefit, so B is NOT a disadvantage.

Yaad rakhein: LAN = fast + cheap to run, but costly to set up and small in reach.

Answer: (B) It has high data transmission rates.

Concept: $n$th term of an A.P.: $a_n=a+(n-1)d$. The difference of two terms gives $d$ directly: $a_m-a_n=(m-n)d$.

1. $a_{28}-a_{15}=416-208=208=13d$ → $d=16$.
2. $a_{49}=a_{28}+(49-28)d=416+21\times 16=416+336=752$.
3. Check via $a$: $a=208-14\times 16=-16$; $a_{49}=-16+48\times 16=752$ ✓.

Trick: Jump directly from a known term: $a_{49}=a_{28}+21d$ – no need to find $a$.

Answer: (C) 752

Concept: When the first term $a$ and last term $l$ are known, $S_n=\frac{n}{2}(a+l)$.

1. $a=20$, $l=55$, $S_n=600$.
2. $600=\frac{n}{2}(20+55)=\frac{75n}{2}$.
3. $n=\frac{1200}{75}=16$.

Trick: Sum = (number of terms) × (average of first and last term). Average $=37.5$, so $n=600÷37.5=16$.

Answer: (B) 16

Concept: Diagonals of a parallelogram bisect each other, so the midpoint of AC equals the midpoint of BD.

1. Midpoint of AC $=(\frac{1+1}{2},\frac{2+0}{2})=(1,1)$.
2. Midpoint of BD $=(\frac{4+k}{2},\frac{3+(-1)}{2})=(\frac{4+k}{2},1)$.
3. Equate $x$-coordinates: $\frac{4+k}{2}=1$ → $4+k=2$ → $k=-2$.

Trick: For parallelogram ABCD (in order): $x_A+x_C=x_B+x_D$ → $1+1=4+k$ → $k=-2$.

Answer: (D) $-2$

Concept: Empirical relation between the three averages: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $87.5=3\times 86-2\text{Mean}=258-2\text{Mean}$.
2. $2\text{Mean}=258-87.5=170.5$.
3. $\text{Mean}=85.25\approx 85.2$ (closest option).

Trick: $\text{Mean}=\frac{3\text{Median}-\text{Mode}}{2}$. Remember "Mode = 3 Med − 2 Mean" as 3M − 2M.

Answer: (A) 85.2

Concept: $a{x}^2+bx+c=0$ has no real roots when the discriminant $D=b^2-4ac<0$.

1. Here coefficients: $1$, $-4a$, $(b+1)$. $D=(-4a{)}^2-4(1)(b+1)=16a^2-4b-4$.
2. No real roots: $16a^2-4b-4<0$ → $4a^2-1<b$.
3. $b>4a^2-1=(2a+1)(2a-1)$.

Trick: $4a^2-1$ is a difference of squares: $(2a{)}^2-1^2=(2a+1)(2a-1)$.

Answer: (C) $b>(2a+1)(2a-1)$

Concept: Painting covers the curved surface; surface area of a sphere $=4\pi r^2$.

1. Diameter $=14$ cm → $r=7$ cm.
2. Surface area of one ball $=4\times \frac{22}{7}\times 7\times 7=616{\text{cm}}^2$.
3. Two balls: $2\times 616=1232{\text{cm}}^2$; cost $=1232\times 3=₹3696$.

Trick: With $r=7$, $4\pi r^2=4\times 22\times 7=616$ instantly (the 7s cancel).

Answer: (D) 3696

Concept: Volume is conserved on recasting; volume of a sphere $=\frac{4}{3}\pi r^3$, so volume $\propto r^3$.

1. $64\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi (12{)}^3$ → $r^3=\frac{1728}{64}=27$.
2. $r=3$ cm.
3. Diameter $=2r=6$ cm.

Trick: $64=4^3$, so each small radius $=\frac{12}{4}=3$ – divide the radius by the cube root of the number of pieces.

Answer: (A) 6

Concept: Angle-sum property: $\angle P+\angle Q+\angle R={180}^{\circ }$. When $3\angle P=4\angle Q=6\angle R$, write each angle in terms of a common constant.

1. Let $3\angle P=4\angle Q=6\angle R=k$ → $\angle P=\frac{k}{3}$, $\angle Q=\frac{k}{4}$, $\angle R=\frac{k}{6}$.
2. Sum: $\frac{k}{3}+\frac{k}{4}+\frac{k}{6}=\frac{4k+3k+2k}{12}=\frac{3k}{4}={180}^{\circ }$ → $k=240$.
3. $\angle P={80}^{\circ }$, $\angle Q={60}^{\circ }$, $\angle R={40}^{\circ }$.
4. $\frac{2\angle P-\angle Q+3\angle R}{5}=\frac{160-60+120}{5}=\frac{220}{5}={44}^{\circ }$.

Trick: $3P=4Q=6R$ → $P:Q:R=\frac{1}{3}:\frac{1}{4}:\frac{1}{6}=4:3:2$ → angles ${80}^{\circ },{60}^{\circ },{40}^{\circ }$.

Answer: (A) ${44}^{\circ }$

Concept: Read only the bars of School C (grey, third bar in each group) and compare their heights.

1. School C: French $=20$, German $=15$, Italian $=35$, Japanese $=30$.
2. The tallest School C bar is in the Italian class ($35$ students).
3. Japanese ($30$) is the tempting second; French and German are much smaller.

Trick: Match the legend colour first, then scan only those bars – ignore the other schools entirely.

Answer: (B) Italian

Concept: $(x+a)(x+b)=x^2+(a+b)x+ab$.

1. Here $a=-9$, $b=8$: $a+b=-1$, $ab=-72$.
2. $(x-9)(x+8)=x^2-x-72$.
3. Check by expanding: $x^2+8x-9x-72=x^2-x-72$ ✓.

Trick: Middle coefficient = sum of the two numbers, constant = their product; sign of the product is negative when the numbers have opposite signs.

Answer: (B) $x^2-x-72$

Concept: Centroid of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

1. $x$-coordinate: $\frac{-5+(-4)+4}{3}=\frac{-5}{3}$.
2. $y$-coordinate: $\frac{7+(-5)+5}{3}=\frac{7}{3}$.
3. $G=(-\frac{5}{3},\frac{7}{3})$.

Trick: Centroid = average of the vertices; the $-4$ and $+4$ cancel in $x$, the $-5$ and $+5$ cancel in $y$.

Answer: (A) $(-\frac{5}{3},\frac{7}{3})$

Concept: In an isosceles triangle the two base angles are equal, and the three angles add to ${180}^{\circ }$.

1. Vertex angle $={76}^{\circ }$, so the two base angles together $={180}^{\circ }-{76}^{\circ }={104}^{\circ }$.
2. Each base angle $=\frac{{104}^{\circ }}{2}={52}^{\circ }$.
3. Options with two unequal angles cannot be base angles of an isosceles triangle.

Trick: Base angle $=\frac{{180}^{\circ }-\text{vertex angle}}{2}$; only the option with equal angles can be right.

Answer: (C) ${52}^{\circ }$ and ${52}^{\circ }$

Concept: Evaluate powers first, then multiplication, then addition/subtraction from left to right (BODMAS).

1. $1^3=1$, $8^2=64$, ${\left(\frac{36}{4}\right)}^2=9^2=81$, $2\times 5=10$.
2. Expression $=1-64+81-0+10$.
3. $1-64=-63$; $-63+81=18$; $18+10=28$.

Trick: Simplify the bracket $\frac{36}{4}=9$ before squaring; it avoids big numbers like $1296$.

Answer: (A) $28$

Concept: Simplify the denominator first, then change division by a fraction into multiplication by its reciprocal.

1. $\frac{2}{3}+\frac{3}{5}=\frac{10+9}{15}=\frac{19}{15}$, so the first part $=\frac{1}{19/15}=\frac{15}{19}$.
2. $\frac{15}{19}÷\frac{8}{39}=\frac{15}{19}\times \frac{39}{8}=\frac{585}{152}$.
3. $\frac{585}{152}=3\frac{129}{152}\approx 3.8487$. Among the options, $3\frac{28}{33}=\frac{127}{33}\approx 3.8485$ is practically equal to this value ($3\frac{16}{19}\approx 3.84$, $3\frac{17}{23}\approx 3.74$, $3\frac{65}{76}\approx 3.86$ are farther).

Trick: $\frac{1}{a/b}=\frac{b}{a}$ – flip the fraction instead of dividing.

Answer: (C) $3\frac{28}{33}$

Concept: $n(T\cup C)=n(T)+n(C)-n(T\cap C)$; neither = total $-n(T\cup C)$.

1. $n(T\cup C)=150+225-100=275$.
2. Neither tea nor coffee $=350-275=75$.
3. Check from the figure: only tea 50, both 100, only coffee 125, neither 75; $50+100+125+75=350$.

Trick: The 100 'both' students are counted twice in $150+225$, so subtract them once.

Answer: (C) $75$

Concept: Work inside the bracket first: union collects all elements, intersection keeps common ones.

1. $B\cup C=\{7,9,11,13,15\}$.
2. $A\cap (B\cup C)$ = elements of $A=\{3,5,7,9,11\}$ that are also in $B\cup C$ = $\{7,9,11\}$.

Trick: Result of $A\cap (\dots )$ can never contain 3 or 5 (absent from B and C) – options B, C, D drop out.

Answer: (A) $\{7,9,11\}$