Q&A: practice questions with answers

Concept: Work backwards: in each step, $\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}$; the last divisor is the HCF.

1. Last step: divisor 57, quotient 8, remainder 0 → dividend $=57\times 8=456$.
2. Middle step: divisor 456, quotient 5, remainder 57 → $456\times 5+57=2337$.
3. First step: divisor 2337, quotient 1, remainder 456 → $2337+456=2793$.
4. $\text{LCM}=\frac{2793\times 2337}{57}=49\times 2337=114513$.

Trick: $\text{LCM}=\frac{a\times b}{\text{HCF}}$; the last digit check $9\times 7=63$ already points to 3, i.e. 114513.

Answer: (D) 114513

Concept: Crossing time $=\frac{\text{sum of lengths}}{\text{relative speed}}$; same direction → subtract speeds, opposite → add.

1. P and Q: relative speed $=30-21=9\text{km/h}=2.5\text{m/s}$; time $=6.5\times 60=390\text{s}$.
2. $P+296=2.5\times 390=975$ → $P=679\text{m}$.
3. P and R (opposite): $30+96=126\text{km/h}=35\text{m/s}$.
4. Time $=\frac{679+196}{35}=\frac{875}{35}=25\text{s}$.

Trick: $\text{km/h}\to \text{m/s}$: multiply by $\frac{5}{18}$; $126\times \frac{5}{18}=35$.

Answer: (C) 25

Concept: Subtract the two given combinations to isolate the cost of tables.

1. $(18c+7t)-(18c+4t)=7700-7100$ → $3t=600$, $t=200$.
2. $18c=7100-4\times 200=6300$.
3. $18c+5t=6300+1000=7300$.

Trick: 18c + 5t = (18c + 4t) + t = 7100 + 200 = 7300.

Answer: (C) ₹7,300

Concept: Monthly saving = salary − expenses; in the birthday month the saving becomes zero.

1. Monthly saving $=13000-4000-2000=7000$.
2. Savings are made in only 11 months of the year.
3. Annual savings $=7000\times 11=77000$.

Trick: 12 months − 1 birthday month = 11 months of saving.

Answer: (B) 77,000

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{gain}\%}{100}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. $\text{SP}=605\times 1.37=828.85$.
2. $\text{MP}\times 0.55=828.85$.
3. $\text{MP}=\frac{828.85}{0.55}=1507$.

Trick: $\text{MP}=605\times \frac{137}{55}=11\times 137=1507$ (since $605=55\times 11$).

Answer: (D) ₹1,507

Concept: Curved surface area of a cone $=\pi rl$ and $h=\sqrt{l^2-r^2}$.

1. Diameter 44 → $r=22\text{cm}$.
2. $\pi \times 22\times l=2684\pi$ → $l=\frac{2684}{22}=122$.
3. $h=\sqrt{{122}^2-{22}^2}=\sqrt{14884-484}=\sqrt{14400}=120$.

Trick: $l^2-r^2=(l-r)(l+r)=100\times 144=14400$, so $h=120$ without big squares.

Answer: (D) 120

Concept: If $DE\parallel AC$, then $△BDE~△BAC$ and the ratio of areas equals the square of the ratio of sides.

1. $BA=BD+DA=17+14=31\text{cm}$.
2. $\frac{ar(△BDE)}{ar(△BAC)}={\left(\frac{17}{31}\right)}^2=\frac{289}{961}$.
3. Trapezium $ADEC=961-289=672$ parts.
4. Required ratio $=289:672$.

Trick: Small triangle : trapezium $=a^2:(b^2-a^2)$ with $a=17,b=31$.

Answer: (B) 289 : 672

Concept: Rule for 7: double the last digit and subtract it from the rest; the result must be divisible by 7.

1. 2436: $243-2\times 6=231$; $23-2=21$ → divisible by 7 ($2436=7\times 348$).
2. 1454: $145-8=137$ ✗; 3366: $336-12=324$ ✗; 3344: $334-8=326$ ✗.

Trick: Repeat the "double and subtract" step until a small number appears.

Answer: (A) 2436

Concept: $a^3+b^3=(a+b{)}^3-3ab(a+b)$.

1. Let $a=5x,b=4y$: $a+b=12$, $ab=20xy=40$.
2. $125x^3+64y^3={12}^3-3\times 40\times 12$.
3. $=1728-1440=288$.

Trick: Spot $125x^3=(5x{)}^3$ and $64y^3=(4y{)}^3$, then use the sum-of-cubes identity.

Answer: (A) 288

Concept: If profit% at $S_1$ equals loss% at $S_2$, then $\text{CP}=\frac{S_1+S_2}{2}$.

1. $\text{CP}=\frac{214+146}{2}=180$ (profit 34 = loss 34, i.e. $18.\bar{8}\%$ each).
2. For 55% profit: $\text{SP}=180\times 1.55=279$.

Trick: Equal profit% and loss% → CP is the average of the two selling prices.

Answer: (C) 279

Concept: $\text{CSA}=\pi rl$, $\text{TSA}=\text{CSA}+\pi r^2$.

1. Diameter 70 → $r=35\text{cm}$.
2. $\pi r^2=\frac{22}{7}\times 35\times 35=3850{\text{cm}}^2$.
3. $\text{TSA}=2970+3850=6820{\text{cm}}^2$.
4. (Slant height $l=\frac{2970}{110}=27\text{cm}$ is not even needed.)

Trick: TSA = CSA + base area; only the base $\pi r^2$ has to be found.

Answer: (A) 6820

Concept: Take total work $=$ LCM of the days and find each person's units per day.

1. Total work $=\text{LCM}(20,45)=180$ units
2. A + B $=\frac{180}{20}=9$ units/day; A $=\frac{180}{45}=4$ units/day
3. B $=9-4=5$ units/day
4. One-third work $=60$ units → $\frac{60}{5}=12$ days

Trick: B alone takes $\frac{20\times 45}{45-20}=36$ days; one-third of it $=12$ days.

Answer: (B) $12$ days

Concept: $\text{SI}=\frac{P\times R\times T}{100}$

1. $650=\frac{1300\times R\times 10}{100}=130R$
2. $R=\frac{650}{130}=5\%$

Trick: Interest is half the principal in 10 years → $\frac{50\%}{10}=5\%$ per year.

Answer: (D) $5\%$

Concept: For $n$ (odd) observations in ascending order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: $17,18,22,28,32,45,57,69,75,79,91$
2. $n=11$ → $\frac{11+1}{2}=6^{\text{th}}$ value
3. $6^{\text{th}}$ value $=45$

Trick: Always sort first – the middle of the unsorted list (57) is a trap.

Answer: (D) $45$

Concept: Successive sales multiply: C's price $=$ A's cost $\times (1-0.60)\times (1+0.25)$.

1. C's price $=x\times 0.4\times 1.25=0.5x$
2. $0.5x=89569$
3. $x=179138$

Trick: $-60\%$ then $+25\%$: $\frac{2}{5}\times \frac{5}{4}=\frac{1}{2}$ – A's cost is double C's price.

Answer: (C) ₹1,79,138

Concept: $(a+b{)}^2$ needs only the value of $a+b$.

1. $a+b=41$
2. $(a+b{)}^2={41}^2=1681$

Trick: Extra data ($a-b=38$) is a distractor; ${41}^2=1600+80+1=1681$.

Answer: (D) $1681$

Concept: A common root of both equations must satisfy the second equation.

1. $x^2-5x+6=0\Rightarrow (x-2)(x-3)=0$, roots $2,3$
2. If $x=2$: $8-2k+6=0\Rightarrow k=7$
3. If $x=3$: $18-3k+6=0\Rightarrow k=8$ (not an option)
4. So $k=7$

Trick: Factorise the easy equation first, then substitute its roots.

Answer: (C) $7$

Concept: Use $(a-b)(a+b)=a^2-b^2$ with numbers around 95.

1. $86\times 104=(95-9)(95+9)$
2. $={95}^2-9^2=9025-81=8944$

Trick: Or $86\times 104=8600+344=8944$; last digit $6\times 4=24$ → ends in 4.

Answer: (A) $8944$

Concept: Write the two time conditions and add them to eliminate Sameer's time.

1. Let Abhay's speed $=a$, Sameer's $=s$. $\frac{162}{a}-\frac{162}{s}=6$
2. Doubled speed: $\frac{162}{s}-\frac{162}{2a}=3$
3. Adding: $\frac{162}{a}-\frac{81}{a}=9\Rightarrow \frac{81}{a}=9$
4. $a=9\text{km/h}$ (Sameer: 12 h; Abhay: 18 h ✓)

Trick: Option check: $a=9$ → 18 h; doubled → 9 h; Sameer $18-6=12$ h and $12-9=3$ ✓.

Answer: (D) $9\text{km/h}$

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference.

1. $AB\parallel DC$, $AB=30$, $DC=86$
2. $EF=\frac{DC-AB}{2}=\frac{86-30}{2}$
3. $EF=28\text{cm}$ (BC and AD are not needed)

Trick: Midpoints of sides → half the sum; midpoints of diagonals → half the difference.

Answer: (A) $28$