Q&A: practice questions with answers

Concept: $\text{Expenditure}=\text{Price}\times \text{Consumption}$, so $\text{new consumption}=\text{new expenditure}÷\text{new price}$.

1. Price factor: $\frac{185}{100}$; required expenditure factor: $\frac{111}{100}$.
2. Consumption factor = $\frac{111}{185}=\frac{3}{5}=0.6$
3. Consumption falls from 1 to 0.6 → reduction = 40%.

Trick: Spot that $111=37\times 3$ and $185=37\times 5$ → ratio 3 : 5 at once, so cut = $\frac{2}{5}=40\%$.

Answer: (C) 40%

Concept: LCM = product of the highest power of every prime present.

1. $180=2^2\times 3^2\times 5$ already covers 30.
2. $234=2\times 3^2\times 13$ also covers 13; it adds only the prime 13.
3. $\text{LCM}=180\times 13=2340$

Trick: Drop numbers that divide another (30 | 180, 13 | 234); then $\text{LCM}(180,234)=180\times 234÷\text{HCF }18=2340$.

Answer: (B) 2340

Concept: With equal rate and time, the interest difference comes only from the principal difference.

1. Principal difference = ₹10200 − ₹5400 = ₹4800
2. SI on ₹4800 for 4 years = ₹720 → $\frac{4800\times 4\times R}{100}=720$
3. $192R=720$ → $R=3.75\%$

Trick: Interest per year on the extra money = $720÷4=$ ₹180; rate = $\frac{180}{4800}\times 100=3.75\%$.

Answer: (A) 3.75

Concept: For 3 years: $\text{CI}-\text{SI}=P\times {\left(\frac{R}{100}\right)}^2\times (3+\frac{R}{100})$.

1. $R=10\%\to {\left(\frac{1}{10}\right)}^2\times (3+0.1)=0.031$
2. $P\times 0.031=93$
3. $P=93÷0.031=$ ₹3000

Trick: At 10%, 3-year difference = 3.1% of P; for 2 years it is 1% of P – memorise both.

Answer: (A) 3000

Concept: For equal distances at speeds x and y, average speed = $\frac{2xy}{x+y}$.

1. $2\times 56\times 72=8064$
2. $56+72=128$
3. $8064÷128=63$ km/h

Trick: Average speed is always a bit below the simple average (64 here) – only 63 fits.

Answer: (D) 63

Concept: Take each prime at its highest power across all numbers.

1. Highest powers: $2^2$ (from 180), $3^2$ (45, 180), $5^2$ (75), 11 (110).
2. $4\times 9=36$; $36\times 25=900$
3. $900\times 11=9900$

Trick: LCM must be divisible by 11 and 25 → it ends in 00 or 25/50/75; only 9900 qualifies.

Answer: (D) 9900

Concept: Same distance both ways → average speed = $\frac{2xy}{x+y}$.

1. $2\times 60\times 30=3600$
2. $60+30=90$
3. $3600÷90=40$ km/h

Trick: Ratio method: speeds 2 : 1 → times 1 : 2; take distance 60 → total 120 km in $1+2=3$ h → 40 km/h.

Answer: (A) 40

Concept: Net effect of +40% mark-up and −25% discount gives the profit %.

1. Net % = $40-25-\frac{40\times 25}{100}=+5\%$
2. 5% of CP = ₹450
3. CP = $450\times 20=$ ₹9000

Trick: Successive % formula $a+b+\frac{ab}{100}$ (use $b=-25$) gives profit % in one line.

Answer: (D) 9,000

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

1. $\text{Area}=4\times \frac{22}{7}\times 49=88\times 7=616{\text{cm}}^2$
2. Rate = ₹$18480÷616$
3. = ₹30 per cm²

Trick: With $r=7$ and $\pi =\frac{22}{7}$, $4\pi r^2=616$ – a value worth remembering.

Answer: (C) ₹30

Concept: For two numbers: $\text{first}\times \text{second}=\text{HCF}\times \text{LCM}$.

1. $16\times \text{other}=8\times 144=1152$
2. Other = $1152÷16=72$
3. Check: $\text{HCF}(16,72)=8$ ✓

Trick: Other number = LCM × HCF ÷ given = $144\times \left(\frac{8}{16}\right)=144÷2=72$ – mental.

Answer: (C) 72

Concept: Set $\frac{a-x}{b-x}$ equal to the new ratio and solve.

1. $(43-x):(91-x)=3:7$
2. Difference of terms stays 48 = 4 parts → 1 part = 12
3. 3 parts = 36 → $x=43-36=7$

Trick: Subtracting the same number keeps the difference fixed: match the difference (48) with the ratio difference (4).

Answer: (C) 7

Concept: Equal amounts after n and n+1 years → shares are in the ratio $(1+\frac{r}{100}):1$.

1. 8% → growth factor $\frac{27}{25}$.
2. Raj (14 yrs) : Varun (15 yrs) = 27 : 25
3. $1612÷52=31$ → Raj = $27\times 31=$ ₹837

Trick: The person whose money grows for fewer years must get the bigger share – rule out answers below half (₹806).

Answer: (B) 837

Concept: Fourth proportional of a, b, c = $\frac{b\times c}{a}$.

1. $\sqrt{24}=2\sqrt{6}$, so $a=10\sqrt{6}$.
2. $b\times c=9\times \sqrt{150}=9\times 5\sqrt{6}=45\sqrt{6}$
3. $d=45\sqrt{6}÷10\sqrt{6}=4.5$

Trick: Simplify all surds to the same root ($\sqrt{6}$) first; they cancel and only plain numbers remain.

Answer: (A) 4.5

Concept: $\text{New member's value}=\text{new average}+(\text{old count}\times \text{rise in average})$.

1. Average rises by 1 year when the teacher joins.
2. Teacher gives 1 year extra to each of 22 students = 22 years.
3. Teacher = $48+22=70$ years

Trick: Avoid big totals: $\text{new value}=\text{new avg}+n\times \text{change}$. Check: $23\times 48-22\times 47=70$.

Answer: (C) 70

Concept: Opposite directions → add speeds; distance covered = sum of lengths.

1. Relative speed = 250 km/h, i.e. $\frac{625}{9}$ m/s.
2. Combined length covered in 5 s ≈ 347.22 m.
3. Removing the known 140 m leaves 207.22 m for the second train.

Trick: Options differ only in decimals – the '.22' from ÷9 ($\frac{2}{9}=0.22$) points straight to 207.22.

Answer: (D) 207.22

Concept: $\text{Product of two numbers}=\text{HCF}\times \text{LCM}$.

1. $\text{HCF}=148176÷3528$
2. $3528\times 40=141120$; remainder $7056=3528\times 2$
3. HCF = 42

Trick: HCF must divide LCM: $3528=42\times 84$, while 51, 32, 36 fail on quick checks.

Answer: (A) 42

Concept: Value after n years = $P\times {(1-\frac{R}{100})}^n$.

1. After year 1: $12500-5\%(625)=11875$
2. After year 2: $11875-5\%(593.75)$
3. = ₹11,281.25

Trick: Two successive −5% = −9.75% net; 9.75% of 12500 = 1218.75 → $12500-1218.75=11281.25$.

Answer: (D) ₹11,281.25

Concept: Find CP from the first sale, then use it for the second.

1. CP = $4824÷1.2=$ ₹4020
2. New profit = $6000-4020=$ ₹1980
3. Profit % = $\frac{1980}{4020}\times 100\approx 49.25\%\approx 49\%$

Trick: 1980 is just under half of 4020 → answer just under 50% → 49%.

Answer: (A) 49%

Concept: If gain on n articles = SP of m articles, then CP of n = SP of (n − m).

1. CP of 153 chairs = SP of $153-78=75$ chairs
2. Profit = SP of 78 chairs on cost of 75 chairs' SP
3. Profit % = $\frac{78}{75}\times 100=104\%$

Trick: Profit % = $\frac{m}{n-m}\times 100$ – one step.

Answer: (A) 104%

Concept: Take total work = LCM of days; the one who leaves works (T − 5) days.

1. Work = 240 units; per day: Ram 15, Ravi 12, Reena 10.
2. Pretend Ravi stays: extra work = $12\times 5=60$ → $240+60=300$
3. All three together do 37/day → T = $\frac{300}{37}=8\frac{4}{37}$ days

Trick: Add the leaver's missing work to the total, then divide by everyone's combined rate.

Answer: (C) $8\frac{4}{37}$ days