11 February 2025 · Shift 2

Time and work, alternate days Take the total work as the LCM of 6 and 20, i.e. 60 units. Sanjay's one day work $=\frac{60}{6}=10$ units; Tannu's one day work $=\frac{60}{20}=3$ units. Tannu begins, so one pair of days (Tannu then Sanjay) finishes 3+10=13 units. In 8 days (4 such pairs) work done $=4\times 13=52$ units. The 9th day is Tannu's: total becomes 52+3=55 units, so 60-55=5 units remain. The 10th day is Sanjay's; he needs $\frac{5}{10}=0.5$ day for the remaining 5 units. Total time =9+0.5=9.5 days, so option (a) is correct.

Profit percentage Let the cost price of one chair be C and its selling price be S. Gain on 154 chairs =154S-154C, and the question says this equals the SP of 66 chairs. 154S-154C=66S $154S-66S=154C\Rightarrow 88S=154C$ $\frac{S}{C}=\frac{154}{88}=\frac{7}{4}$, so take C=4 and S=7. Profit % $=\frac{S-C}{C}\times 100=\frac{7-4}{4}\times 100=75\%$ So option (b) is correct.

Simple interest, principal Formula: $SI=\frac{P\times R\times T}{100}$ Here SI=387, R=6 and T=2. $387=\frac{P\times 6\times 2}{100}=\frac{12P}{100}$ $12P=387\times 100=38700$ $P=\frac{38700}{12}=3225$ So the principal is ₹3,225 and option (b) is correct.

BODMAS simplification Apply BODMAS, opening the innermost bracket first. 13-11=2, so $\frac{13-11}{2}=\frac{2}{2}=1$ Inside the braces: $5÷5\times 1=1\times 1=1$ $16\times 12=192$ $192\times 1=192$ So option (a) is correct.

Weighted average Weighted average = (total marks of all the students) ÷ (total number of students). Group total marks = number of students × their average. Total marks $=20\times 40+25\times 34+15\times 26=800+850+390=2040$ Total students =20+25+15=60 Weighted average $=\frac{2040}{60}=34$ So option (c) is correct. (The 50-mark maximum is extra information and is not needed.)

Average speed, return journey P to Q: 45 minutes $=\frac{45}{60}=\frac{3}{4}$ hour. Speed on the onward trip $=\frac{60}{3/4}=60\times \frac{4}{3}=80$ km/hr. The two trips cover the same distance, so use $\mathrm{average}\mathrm{speed}=\frac{2xy}{x+y}$. $=\frac{2\times 80\times 40}{80+40}=\frac{6400}{120}$ =53.33 km/hr So option (d) is correct. (Averaging 80 and 40 to get 60 is the trap - time, not distance, differs.)

CI minus SI, three years For 3 years, $CI-SI=P\left({\left(1+\frac{R}{100}\right)}^3-1-\frac{3R}{100}\right)$ With R=10: ${\left(1+\frac{10}{100}\right)}^3={\left(1.1\right)}^3=1.331$ $CI-SI=P\left(1.331-1-0.3\right)=0.031P$ 0.031P=124 $P=\frac{124}{0.031}=\frac{124000}{31}=4000$ So the principal is ₹4,000 and option (a) is correct.

Effective discount "Buy five, get one free" means the customer takes 6 articles but pays for only 5. Let the marked price of one article be ₹1, so the marked price of 6 articles is ₹6. Amount actually paid = ₹5, so the discount is ₹1 on a marked price of ₹6. Discount % $=\frac{1}{6}\times 100=16.67\%$ So option (d) is correct. (Dividing by 5 instead of 6 gives 20%, the usual slip.)

Average speed of a journey Let the whole journey be 5 units of distance, so the two parts are 3 units and 2 units. Time for the first part $=\frac{3}{84}=\frac{1}{28}$ hr Time for the remaining part $=\frac{2}{56}=\frac{1}{28}$ hr Total time $=\frac{1}{28}+\frac{1}{28}=\frac{2}{28}=\frac{1}{14}$ hr Average speed $=\frac{\mathrm{total}\mathrm{distance}}{\mathrm{total}\mathrm{time}}=\frac{5}{1/14}=5\times 14=70$ km/hr So option (a) is correct.

Volume of a cone First make the units match: height =0.12 m $=0.12\times 100=12$ cm. Formula: $V=\frac{1}{3}\pi r^{2}h$ $168\pi =\frac{1}{3}\pi r^2\times 12$ $168\pi =4\pi r^2$ $r^2=\frac{168}{4}=42$ $r=\sqrt{42}$ cm, so option (c) is correct.

Compound interest, equal amounts Let Ajay get A and Prakash get P, with A+P=1218. Their amounts are equal, Ajay's after 14 years and Prakash's after 15 years: $A{\left(1+\frac{10}{100}\right)}^{14}=P{\left(1+\frac{10}{100}\right)}^{15}$ Cancel ${\left(\frac{11}{10}\right)}^{14}$ from both sides: $A=P\times \frac{11}{10}$ So A:P=11:10; total =11+10=21 units. 21 units $=1218\Rightarrow 1$ unit $=\frac{1218}{21}=58$ Prakash's share $=10\times 58=580$, i.e. ₹580, so option (b) is correct.

Gain percentage Let the cost price of one toy be C and the selling price be S. 30S=50C $\frac{S}{C}=\frac{50}{30}=\frac{5}{3}$, so take C=3 and S=5. Gain on one toy =5-3=2. Gain % $=\frac{2}{3}\times 100=66.67\%$ So option (c) is correct. (Dividing the gain by the SP gives 40%, which is not asked.)

Ratio of marks Maths $=2\times$ Science, so Maths : Science =2:1. Maths : Computer =4:3. Scale the first ratio so that Maths becomes 4: Maths : Science =4:2. Combining on the common term Maths, we get Maths : Science : Computer =4:2:3. Total =4+2+3=9 units =234 1 unit $=\frac{234}{9}=26$ Marks in maths $=4\times 26=104$, so option (b) is correct.

BODMAS simplification By BODMAS, division and multiplication come first, taken from left to right, and subtraction last. $32÷8=4$ $4\times 5=20$ $2\times 3=6$ 20-6=14 So option (c) is correct.

Ratio of ages Let the present ages of A and B be 4x and 9x years. $9x-4x=10\Rightarrow 5x=10\Rightarrow x=2$ Present ages $=4\times 2=8$ years and $9\times 2=18$ years. After 4 years: 8+4=12 years and 18+4=22 years. Required ratio =12:22=6:11 So option (c) is correct. (12 : 22 must be reduced - leaving it unreduced is the trap.)

Speed and time, inverse ratio For a fixed distance, time is inversely proportional to speed. Speed becomes $\frac{1}{4}$ of normal, so the time becomes 4 times the normal time. Let the normal time be 1 unit; then the slow time is 4 units. Extra time =4-1=3 units, and this delay is 45 minutes. 1 unit $=\frac{45}{3}=15$ minutes Normal time =1 unit =15 minutes, so option (c) is correct.

Percentage, finding the whole "2% of x" means $\frac{2}{100}\times x$. $\frac{2}{100}\times x=336$ $x=336\times \frac{100}{2}=336\times 50$ x=16800 So option (c) is correct. (Multiplying by 100 instead of 50 gives 33600, the trap option.)

Successive percentage change Take the original price as 100 and apply each change to the price standing at that moment. A fall of $r\%$ multiplies the price by $\frac{100-r}{100}$; a rise of $r\%$ multiplies it by $\frac{100+r}{100}$. Final price $=100\times \frac{75}{100}\times \frac{90}{100}\times \frac{70}{100}\times \frac{160}{100}$ $=100\times 0.75\times 0.9\times 0.7\times 1.6=75.6$ Net change =75.6-100=-24.4, i.e. a fall of $\frac{24.4}{100}\times 100=24.4\%$. So the price decreases by 24.4% and option (c) is correct.

HCF and LCM relation For any two numbers: HCF × LCM = product of the two numbers. Let the LCM be L. $24\times L=5184$ $L=\frac{5184}{24}=216$ So option (d) is correct. (A quick check: the LCM must be a multiple of the HCF, and $216=24\times 9$.)

BODMAS simplification Apply BODMAS, innermost bracket first. $65÷5=13$ $\frac{16}{8}=2$ and $\frac{14}{2}=7$ (6-4)=2, so $7\times 2=14$ Inside the braces: 2+14=16 (multiplication before addition). $13\times 16=208$ So option (b) is correct.