21 February 2025 · Shift 3

Simple interest rate Formula: $SI=\frac{P\times R\times T}{100}$. The rate and the time (6 years) are the same for both men, so the extra ₹360 of interest comes only from the extra principal. Extra principal =9400-5400=4000 (in ₹). So $\frac{4000\times R\times 6}{100}=360$. $240R=360\Rightarrow R=\frac{360}{240}=1.5$. Hence the rate of interest is 1.5% per annum, option (d).

LCM by prime factors Break each number into primes: $26=2\times 13$, $34=2\times 17$, $221=13\times 17$, $195=3\times 5\times 13$. The LCM takes every prime that appears, at its highest power - here each of 2,3,5,13,17 appears only to the first power. $LCM=2\times 3\times 5\times 13\times 17$. $=30\times 221=6630$. A quick check: 6630 is the only option divisible by 17, so the correct option is (d).

Average of numbers Average = (sum of the numbers) ÷ (how many numbers); here there are 6 numbers. Sum =309+125+90+45+125+104. 309+125=434, 434+90=524, 524+45=569, 569+125=694, 694+104=798. Average $=\frac{798}{6}=133$. Hence the correct option is (d).

Circular track meeting To meet at the STARTING POINT each runner must have finished a whole number of laps, so only the individual lap times matter - the direction of running does not affect this answer. Convert the speeds: $27\times \frac{5}{18}=7.5$ m/s and $45\times \frac{5}{18}=12.5$ m/s. A's lap time $=\frac{1500}{7.5}=200$ s. B's lap time $=\frac{1500}{12.5}=120$ s. Both are back at the start together after LCM(200,120)=600 s. 600 s $=\frac{600}{60}=10$ minutes, so the correct option is (a).

Volume of a hemisphere Inner radius $=\frac{1.4}{2}=0.7$ m. Capacity of a hemisphere $=\frac{2}{3}\pi r^3$. $=\frac{2}{3}\times \frac{22}{7}\times {\left(0.7\right)}^3$ $=\frac{2}{3}\times \frac{22}{7}\times 0.343=\frac{2}{3}\times 1.078$ $=0.71866\dots \approx 0.72$ Hence the capacity is $0.72m^3$, option (b).

Compound interest shares Formula: $A=P{\left(1+\frac{R}{100}\right)}^n$. Anil's amount after 11 years equals Ashok's after 12 years: $P_1{\left(\frac{108}{100}\right)}^{11}=P_2{\left(\frac{108}{100}\right)}^{12}$. Cancelling ${\left(\frac{108}{100}\right)}^{11}$ from both sides gives $\frac{P_1}{P_2}=\frac{108}{100}=\frac{27}{25}$. The son with the SHORTER time must get the larger share, which matches 27>25. Total parts =27+25=52, so one part $=\frac{1612}{52}=31$. Anil's share $=27\times 31=837$. Hence Anand gave ₹837 to Anil, option (a).

Equivalent discount A customer pays for 7 notebooks and carries home 7+3=10 notebooks. So 10 notebooks are paid at the marked price of only 7: the money saved is on 3 out of every 10. Equivalent discount $=\frac{3}{10}\times 100=30\%$. Sale price $=185\times \left(1-\frac{30}{100}\right)=185\times \frac{7}{10}$. $=\frac{1295}{10}=129.5$. Hence the sale price is ₹129.50, option (a).

Average and totals Turn every average into a total: total = average × number of people. Tushar + Salil + Mukund $=3\times 46=138$ kg. Tushar + Salil $=2\times 40=80$ kg. So Mukund =138-80=58 kg. Salil + Mukund $=2\times 46=92$ kg. Salil =92-58=34 kg. Hence the correct option is (a).

Fourth proportional If a:b::c:d, then the product of the extremes equals the product of the means: $a\times d=b\times c$. So the fourth proportional is $d=\frac{b\times c}{a}$, with a=20, b=24, c=30. $d=\frac{24\times 30}{20}=\frac{720}{20}=36$. Hence the fourth proportional is 36, option (a).

BODMAS simplification BODMAS: clear the innermost bracket first, then the braces, then the square bracket. Innermost bracket: 9-6=3. Inside the braces: $\frac{52}{2}=26$ and $\frac{18}{3}\times 3=6\times 3=18$. So the braces give 26+18=44. The round bracket gives $87÷3=29$. Value $=29\times 44=1276$. Hence the correct option is (c).

Profit and selling price ₹1250 per hundred means 400 oranges are 4 hundreds. Cost price of 400 oranges $=1250\times 4=5000$ (in ₹). Selling price of 400 oranges =5000+1000=6000 (in ₹). Selling price of one orange $=\frac{6000}{400}=15$. Selling price per dozen $=15\times 12=180$. Hence the selling price per dozen is ₹180, option (a).

Selling price at a profit Formula: $SP=CP\times \frac{100+\mathrm{Profit}\%}{100}$. $SP=500\times \frac{100+22}{100}=500\times \frac{122}{100}$. $=5\times 122=610$. Hence he should sell it for ₹610, option (c).

Ratio to actual number Let the numbers of male and female employees be 4x and 3x. Given 4x=460, so $x=\frac{460}{4}=115$. Female employees $=3x=3\times 115=345$. Hence the correct option is (b).

Time and work efficiency Take the whole work as 100%. All three together did 46% in 4 days, so their combined one-day work $=\frac{46}{4}=11.5\%$. The rest, $100-46=54\%$, was done by Pinki and Chinki in 9 days, so their one-day work $=\frac{54}{9}=6\%$. Reema's one-day work $=11.5\%-6\%=5.5\%$. Reema alone needs $\frac{100}{5.5}=\frac{1000}{55}=\frac{200}{11}$ days. $\frac{200}{11}=18\frac{2}{11}$ days, so the correct option is (a).

Trains crossing, same direction To cross completely, the faster train must cover BOTH lengths: 410+440=850 m. Same direction, so the relative speed is the difference: 70-60=10 km/h. 10 km/h $=10\times \frac{5}{18}=\frac{25}{9}$ m/s. Time = distance ÷ relative speed $=850÷\frac{25}{9}=850\times \frac{9}{25}$. $=34\times 9=306$ seconds. In minutes, $\frac{306}{60}=5.1$, so the correct option is (c).

Percentage increase The rise is $15\frac{3}{4}\%=\frac{63}{4}\%$. Increase $=18000\times \frac{63}{4\times 100}=18000\times \frac{63}{400}$. $=45\times 63=2835$ (in ₹). Current price =18000+2835=20835. Hence the current price is ₹20,835, option (d).

Compound interest shares Formula: $A=P{\left(1+\frac{R}{100}\right)}^n$, with R=10 so the yearly factor is $\frac{11}{10}$. Anil's amount after 15 years equals Ashok's after 16 years: $P_{Anil}{\left(\frac{11}{10}\right)}^{15}=P_{Ashok}{\left(\frac{11}{10}\right)}^{16}$. Cancelling ${\left(\frac{11}{10}\right)}^{15}$ gives $\frac{P_{Anil}}{P_{Ashok}}=\frac{11}{10}$. Total parts =11+10=21, so one part $=\frac{1218}{21}=58$. Ashok's share $=10\times 58=580$. Hence Anand gave ₹580 to Ashok, option (a).

LCM from a ratio Let the numbers be 5x and 3x; since 5 and 3 have no common factor, x is their HCF. For such a pair, $LCM=5\times 3\times x=15x$. $15x=105\Rightarrow x=\frac{105}{15}=7$. The numbers are $5\times 7=35$ and $3\times 7=21$. Sum =35+21=56. Hence the correct option is (d).

Price, consumption, expenditure Expenditure = price $\times$ consumption, so fix convenient starting values. Let the price be 100 per litre and the consumption 100 litres, so the expenditure is $100\times 100=10000$. New price $=100\times \frac{175}{100}=175$; new expenditure $=10000\times \frac{126}{100}=12600$. New consumption $=\frac{12600}{175}=72$ litres. Fall in consumption =100-72=28 out of 100, i.e. $\frac{28}{100}\times 100=28\%$. Hence the consumption must be cut by 28%, option (d).

Integers with BODMAS BODMAS order: division first, then multiplication, then addition and subtraction. $\left(-60\right)÷\left(-12\right)=5$ (a negative divided by a negative is positive). $\left(-4\right)\times 6=-24$. The expression becomes (-9)-5+(-24). =-9-5-24=-38. Hence the correct option is (d).