12 February 2025 · Shift 2

Successive percentage change Percentages of different bases never add - apply each change to the price standing at that moment, so multiply the factors. A fall of $a\%$ multiplies the price by $\left(1-\frac{a}{100}\right)$ and a rise of $b\%$ multiplies it by $\left(1+\frac{b}{100}\right)$. Take the original price as ₹100. Net multiplier $=0.75\times 0.85\times 0.55\times 1.65$ $0.75\times 0.85=0.6375$ $0.6375\times 0.55=0.350625$ $0.350625\times 1.65=0.57853125$ So ₹100 becomes ₹57.853, a change of 57.853-100=-42.147 Net change $=-42.14\%$ (correct to two decimals), i.e. a decrease of about $42.14\%$. Hence the price decreases by $42.14\%$, so the correct option is (c).

Simple and compound interest Formula: $CI=P\left({\left(1+\frac{R}{100}\right)}^2-1\right)$ and $SI=\frac{P\times R\times T}{100}$, the principal being the same in both. ${\left(1+\frac{4}{100}\right)}^2={\left(1.04\right)}^2=1.0816$, so two years of compounding add $8.16\%$ of the principal. Two years of simple interest add $2\times 4=8\%$ of the principal. From the compound interest: $8.16\%$ of P=102 $P=\frac{102\times 100}{8.16}=1250$, so the sum invested is ₹1,250. $SI=\frac{1250\times 4\times 2}{100}=100$ Shortcut worth keeping: the principal is common, so $SI=CI\times \frac{8}{8.16}=102\times \frac{8}{8.16}=100$. Hence the corresponding simple interest is ₹100, so the correct option is (c).

Volume of a sphere Formula: volume of a sphere $V=\frac{4}{3}\pi r^3$ (a sphere, not a hemisphere, so no halving). Substitute r=10 cm and $\pi =3.141$: $V=\frac{4}{3}\times 3.141\times {10}^3$ $=\frac{4}{3}\times 3.141\times 1000=\frac{12564}{3}$ =4188 Hence the volume is $4188{\mathrm{cm}}^3$, so the correct option is (a).

Percentage of a number '$2\%$ of x' means $\frac{2}{100}\times x$ - per cent is simply 'out of a hundred'. $\frac{2}{100}\times x=264$ $x=\frac{264\times 100}{2}$ $x=132\times 100=13200$ Check: $2\%$ of 13200=264. (Option (a) 26400 is what you get by dividing by 1 per cent instead of 2.) Hence x=13200, so the correct option is (c).

Trains crossing each other When two trains cross each other the distance covered is the sum of their lengths, not either length alone. Total distance =250+350=600 m Moving in opposite directions the speeds add: relative speed =35+25=60 km/h. Convert to metres per second: $60\times \frac{5}{18}=\frac{50}{3}$ m/s Time $=\frac{\mathrm{total}\mathrm{length}}{\mathrm{relative}\mathrm{speed}}=\frac{600}{\frac{50}{3}}=600\times \frac{3}{50}$ =36 seconds Hence they cross in 36 seconds, so the correct option is (b).

Time and work If $\frac{1}{12}$ of the work takes 9 days, the whole work takes $12\times 9=108$ days. The work still left $=1-\frac{1}{12}=\frac{11}{12}$ Days still needed $=\frac{11}{12}\times 108=11\times 9=99$ The question asks how many MORE days, so 108 (option b) is the trap - that is the total time. Hence he needs 99 more days, so the correct option is (d).

Weighted average The combined average is total amount ÷ total members; the plain average of ₹375, ₹309 and ₹298 would be wrong because the groups are of different sizes. Group A: $38\times 375=14250$ Group B: $31\times 309=9579$ Group C: $52\times 298=15496$ Total amount =14250+9579+15496=39325 Total members =38+31+52=121 Average $=\frac{39325}{121}=325$ Hence ₹325 is spent per member, so the correct option is (c).

Speed, time and distance Over a fixed distance, time varies inversely with speed: $\mathrm{time}\propto \frac{1}{\mathrm{speed}}$. Speed becomes $\frac{3}{8}$ of normal, so the time becomes $\frac{8}{3}$ of normal - speed and time ratios are inverses, 8:3 and 3:8. Let the normal time be t minutes, so the slower time is $\frac{8}{3}t$ minutes. Extra time $=\frac{8}{3}t-t=\frac{5}{3}t$ $\frac{5}{3}t=45$ $t=45\times \frac{3}{5}=27$ The trap is to give the late arrival's time $\frac{8}{3}\times 27=72$; the question asks for the normal time. Hence he would have taken 27 minutes, so the correct option is (d).

Successive discounts Successive discounts multiply the fraction of the price that survives; they never add (16 + 10 + 25 = 51% is the standard wrong answer). Fraction paid $=\left(1-0.16\right)\left(1-0.10\right)\left(1-0.25\right)=0.84\times 0.90\times 0.75$ $0.84\times 0.90=0.756$ $0.756\times 0.75=0.567$ So $56.7\%$ of the marked price is paid. Single equivalent discount $=100\%-56.7\%=43.3\%$ Hence the equivalent single discount is $43.3\%$, so the correct option is (b).

Profit and selling price Cost price of 100 guavas is ₹1230, and 400 guavas are four such hundreds. $CP=1230\times 4=4920$ Selling price = cost price + profit =4920+880=5800 Selling price of one guava $=\frac{5800}{400}=14.5$ Selling price of a dozen $=14.5\times 12=174$ Hence the selling price is ₹174 per dozen, so the correct option is (c).

Simple interest rate Formula: $SI=\frac{P\times R\times T}{100}$. The rate and the time are the same for both men, so the extra interest comes only from the extra principal. Extra principal =9400-5400=4000 $\frac{4000\times R\times 4}{100}=840$ 160R=840 $R=\frac{840}{160}=5.25$ Hence the rate is $5.25\%$ per annum, so the correct option is (d).

LCM based number 'Decreased by 3 and then divisible by 15, 25 and 55' means the number is a common multiple of the three, plus 3. $15=3\times 5$, $25=5\times 5$, $55=5\times 11$ Take the highest power of each prime: $\mathrm{LCM}=3\times 5^2\times 11=825$ Smallest such number =825+3=828 Check: 828-3=825, and 825 is divisible by 15, 25 and 55. Option (b) 825 is the LCM itself, not the number asked for. Hence the number is 828, so the correct option is (a).

Simplification of brackets Peel the brackets from the inside outwards - round $\left(\right)$, then curly $\left\{\right\}$, then square $\left(\right)$ - using BODMAS inside each. Innermost round bracket: 15-13=2, so $\frac{15-13}{2}=\frac{2}{2}=1$ Curly bracket, division before multiplication: $3÷3\times 1=1\times 1=1$ The other round bracket: $27\times 5=135$ Square bracket: $135\times 1=135$ Hence the value is 135, so the correct option is (b).

Pipes and cisterns The tank is fixed work, so pipes $\times$ time is constant: $n_1{t}_1=n_2{t}_2$. 1 hour 45 minutes =105 minutes. $6\times 105=3\times t$ $t=\frac{630}{3}=210$ minutes 210 minutes =3 hours 30 minutes (halving the pipes simply doubles the time). Hence it takes 3 h 30 min, so the correct option is (d).

Simplification of brackets Work from the innermost bracket outwards, applying BODMAS at every stage. Innermost round bracket: 18-14=4, so $\frac{18-14}{2}=\frac{4}{2}=2$ Curly bracket, division before multiplication: $6÷3\times 2=2\times 2=4$ The other round bracket: $17\times 9=153$ Square bracket: $153\times 4=612$ Hence the value is 612, so the correct option is (b).

Average of a part group Averages cannot be added or subtracted directly - turn each one into a sum first, using sum = average $\times$ count. Sum of all 12 numbers $=12\times 49.5=594$ Sum of the first 5 $=5\times 45=225$ Sum of the next 4 $=4\times 53.25=213$ Sum of the remaining 12-5-4=3 numbers =594-225-213=156 Average of those 3 $=\frac{156}{3}=52$ Hence the average of the remaining three numbers is 52, so the correct option is (b).

HCF with remainders If a divisor leaves remainder r in N, then it divides N-r exactly, so subtract each remainder first. 175-4=171 and 385-5=380 The greatest such divisor is $\mathrm{HCF}\left(171,380\right)$. $171=3\times 3\times 19$ and $380=2\times 2\times 5\times 19$ $\mathrm{HCF}=19$ A divisor must be larger than every remainder it leaves; 19>5 and 19>4, so 19 is valid, while 21 and 22 do not divide 171. Hence the greatest number is 19, so the correct option is (d).

Time and work with men leaving Measure the job in man-days: total work = men $\times$ days $=20\times 15=300$ man-days. Work done in the first 6 days $=20\times 6=120$ man-days Work still left =300-120=180 man-days Workers left =20-10=10 Additional days $=\frac{180}{10}=18$ Hence 18 more days are needed, so the correct option is (a).

Difference of CI and SI For 2 years the whole difference between compound and simple interest is one year's interest on the FIRST year's interest. CI-SI=60-40=20 Simple interest for one year $=\frac{40}{2}=20$, and ₹20 is exactly the interest earned on that ₹20. $\frac{20\times R\times 1}{100}=20$ R=100 per cent per annum. Now put this back into the simple interest: $\frac{P\times 100\times 2}{100}=40$ 2P=40, so P=20 Check: at $100\%$, ₹20 grows to ₹40 then ₹80, so CI=80-20=60 and SI=40 - both match. Hence the principal is ₹20, so the correct option is (d).

Profit percentage Put both purchases together before comparing with the sale. Total mangoes =15+9=24, which is exactly 2 dozen. Total cost price =250+200=450 Total selling price $=240\times 2=480$ Profit =480-450=30 Profit $\%=\frac{\mathrm{profit}}{\mathrm{cost}\mathrm{price}}\times 100=\frac{30}{450}\times 100=\frac{100}{15}$ $=6.67\%$ (to two decimal places) Dividing the profit by the selling price instead would give $6.25\%$, and reading 480 as the profit gives the $106.67\%$ trap. Hence the profit is $6.67\%$, so the correct option is (b).