Q&A: practice questions with answers

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{profit}\%}{100}$.

1. Profit = 30% of ₹520 = ₹156
2. SP = $520+156=$ ₹676

Trick: 30% profit → multiply CP by 1.3: $520\times 1.3=676$.

Answer: (C) 676

Concept: Solve the innermost bracket first, then ÷ and × from left to right.

1. $\frac{15-12}{3}=\frac{3}{3}=1$
2. $\{6÷6\times 1\}=1\times 1=1$
3. $(12\times 10)\times 1=120$

Trick: If the curly bracket reduces to 1, the answer is simply $12\times 10$.

Answer: (A) 120

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

1. Take price = 100, consumption = 100 → expenditure = 10000.
2. New price = 175, new expenditure = 14700.
3. New consumption = $14700÷175=84$ → cut = $100-84=16\%$.

Trick: Consumption factor = $1.47÷1.75=0.84$ → reduce by 16%.

Answer: (A) 16%

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

1. 25% of 45,880 = 11,470
2. Price paid = $45,880-11,470=$ ₹34,410

Trick: 25% off = pay $\frac{3}{4}$: $45,880÷4=11,470$, × 3 = 34,410.

Answer: (B) 34,410

Concept: $\text{Perimeter}=2(l+b)$; $\text{Area}=l\times b$.

1. $l+b=26÷2=13$
2. $b=13-7=6$ cm
3. Area = $7\times 6=42{\text{cm}}^2$

Trick: Half the perimeter gives l + b directly.

Answer: (B) 42

Concept: Greatest number leaving the same remainder = HCF of the differences of the numbers.

1. Differences: $84-42=42$, $153-84=69$, $153-42=111$
2. $\text{HCF}(42,69)=3$, and $111÷3=37$ exactly
3. So the greatest such number is 3

Trick: Check options from the largest: 9 and 7 do not divide 69; 3 does.

Answer: (C) 3

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

1. Ashok (13 yrs) : Raj (14 yrs) = 1.04 : 1 = 26 : 25
2. Total parts = 51 → 1 part = $1224÷51=24$
3. Raj = $25\times 24=$ ₹600

Trick: One-year gap at 4% → ratio 104 : 100 = 26 : 25; the longer-invested person gets the smaller share.

Answer: (C) 600

Concept: Take total work = LCM; find work done in one 2-day cycle.

1. Work = $\text{LCM}(15,24)=120$ units; A = 8/day, B = 5/day.
2. One cycle (2 days) = 13 units; 9 cycles = 18 days → 117 units.
3. Left 3 units, A's turn: $\frac{3}{8}$ day → total $18\frac{3}{8}$ days.

Trick: Stop the cycles just before the total is reached, then give the leftover to whoever's turn it is.

Answer: (B) $18\frac{3}{8}$

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

1. $P=284\times 100÷(5\times 2)$
2. = $28400÷10=$ ₹2840

Trick: 5% for 2 years = 10% of P → $P=284\times 10$.

Answer: (B) 2840

Concept: Common member = (sum of the two pair totals) − (total of all three).

1. $A+B+R=46\times 3=138$
2. $A+B=39\times 2=78$; $B+R=50\times 2=100$
3. Baban = $78+100-138=40$ kg

Trick: Baban is counted twice in the pair totals – subtract the trio total once.

Answer: (D) 40

Concept: Same direction: time to meet = track length ÷ (difference of speeds).

1. Relative speed = $12-9=3$ m/s
2. Time = $432÷3=144$ seconds

Trick: Same direction → subtract speeds; opposite direction → add.

Answer: (B) 144 seconds

Concept: Innermost bracket first; then ÷ and × in left-to-right order.

1. $\frac{18-15}{3}=1$
2. $\{8÷2\times 1\}=4$
3. $(14\times 9)\times 4=126\times 4=504$

Trick: Unit-digit check: $6\times 4$ ends in 4 – only 504 qualifies.

Answer: (B) 504

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

1. Arjun (12 yrs) : Kartik (13 yrs) = 1.1 : 1 = 11 : 10
2. 21 parts = 1218 → 1 part = 58
3. Kartik = $10\times 58=$ ₹580

Trick: At 10%, a 1-year gap gives 11 : 10 – the one investing longer gets 10 parts.

Answer: (C) 580

Concept: $\text{Required decrease}\%=\frac{\text{increase}}{\text{new value}}\times 100$.

1. 100 → 150 → 225
2. Reduce 125 out of 225
3. $\frac{125}{225}\times 100=\frac{500}{9}=55\frac{5}{9}\%$

Trick: Two +50% = $\times \frac{9}{4}$; to undo multiply by $\frac{4}{9}$ → fall of $\frac{5}{9}=55\frac{5}{9}\%$.

Answer: (B) 55 5/9%

Concept: Mean proportional of a and b = $\sqrt{ab}$.

1. $6\times 150=900$
2. $\sqrt{900}=30$

Trick: It is the geometric mean – it must lie between 6 and 150 and satisfy 6 : x = x : 150.

Answer: (D) 30

Concept: Find total SP, then price per item, then multiply by 12.

1. CP of 400 = $4\times 1210=$ ₹4840
2. SP = 4840 + 960 = ₹5800 → per mango ₹14.50
3. Per dozen = $14.5\times 12=$ ₹174

Trick: 400 mangoes = $\frac{100}{3}$ dozen → $5800÷\left(\frac{100}{3}\right)=174$.

Answer: (B) 174

Concept: Third proportional of a and b = $\frac{b^2}{a}$.

1. For 12, 42: ${42}^2÷12=1764÷12=147$
2. For 16, 56: ${56}^2÷16=3136÷16=196$
3. 147 : 196 = 3 : 4 (÷ 49)

Trick: Both pairs have ratio 2 : 7; third proportionals then scale like the first terms (12 : 16 = 3 : 4).

Answer: (A) 3 : 4

Concept: $\text{Average}=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Total = $(2.2+1.8)+(1.6+2.4)+2=4+4+2=10$
2. Average = $10÷5=2$ s

Trick: Pair them: $2.2+1.8=4$, $1.6+2.4=4$ – values balance around 2.

Answer: (C) 2

Concept: Overtaking: distance = sum of lengths; speed = difference of speeds.

1. Distance = $300+200=500$ m = 0.5 km
2. Relative speed = $120-70=50$ km/h
3. Time = $0.5÷50$ h = $\frac{1}{100}$ h = 0.6 min

Trick: 1 h = 60 min, so $\frac{1}{100}$ h = $\frac{60}{100}=0.6$ min.

Answer: (D) 0.6

Concept: Largest divisor giving equal remainders = HCF of the differences.

1. $7635-7516=119$; $7992-7635=357$
2. $357=3\times 119$, so HCF(119, 357) = 119 → x = 119
3. Product of digits = $1\times 1\times 9=9$

Trick: Two differences are enough – the third is their sum.

Answer: (A) 9