20 February 2025 · Shift 1

Averages and totals Change every average into a total, because totals can be added and subtracted while averages cannot: total = average $\times$ number of persons. $M+S+C=46\times 3=138$ $M+S=39\times 2=78$ $S+C=45\times 2=90$ Adding the last two totals counts Sachin twice: (M+S)+(S+C)=78+90=168, that is (M+S+C)+S=168. 138+S=168 S=168-138=30 Sachin weighs 30 kg, so the correct option is (d).

LCM by prime factorisation For an LCM, break every number into primes and then keep the highest power of each prime that appears anywhere. $42=2\times 3\times 7$ $36=2^2\times 3^2$ $312=2^3\times 3\times 13$ $126=2\times 3^2\times 7$ The highest powers available are $2^3$, $3^2$, 7 and 13. LCM $=8\times 9\times 7\times 13=504\times 13=6552$ So the correct option is (d).

Average speed Average speed is total distance divided by total time; it is never the average of the two separate speeds. Second stretch $=2\times 30=60$ km, covered in 60 minutes. Total distance =30+60=90 km Total time =45+60=105 minutes $=\frac{105}{60}=\frac{7}{4}$ hours Average speed $=90÷\frac{7}{4}=90\times \frac{4}{7}=\frac{360}{7}$ $=51.4285\dots \approx 51.43$ km/hr So the correct option is (a).

LCM with a common remainder A number that leaves the same remainder 9 with 12, 16 and 24 must be (a multiple of their LCM) +9. $12=2^2\times 3$, $16=2^4$, $24=2^3\times 3$, so LCM $=2^4\times 3=48$. The number has to stay below 5500, so the multiple of 48 must be below 5500-9=5491. $5491÷48=114$ with remainder 19, so the largest usable multiple is $48\times 114=5472$. Required number =5472+9=5481 5481 lies between 5000 and 5500 and leaves 9 with each divisor, so the correct option is (d).

Marked price, discount and profit Take the cost price as ₹100. Note the two percentages sit on different bases: the 30% is on the cost price, the 10% discount is on the marked price. Marked price $=100\times \frac{130}{100}=130$ Selling price $=130\times \frac{90}{100}=117$ Profit =117-100=17 Profit $\%=\frac{17}{100}\times 100=17\%$ So the correct option is (d).

BODMAS order of operations BODMAS: every division and multiplication is cleared first, taken from left to right, and only after that come addition and subtraction. Expression: $32÷8\times 3-3\times 4$ Left to right, division comes first: $32÷8=4$, then $4\times 3=12$. The other product is $3\times 4=12$. 12-12=0 So the value is 0 and the correct option is (a).

Successive profit and loss Each sale is worked on the previous owner's own cost price, so the two changes multiply one after the other; they never simply add up. Let Kanchan's cost price be x. Mohan pays $x\times \frac{105}{100}$, and Pramod pays that amount $\times \frac{75}{100}$. $x\times \frac{105}{100}\times \frac{75}{100}=3150$ $x\times \frac{63}{80}=3150$ $x=3150\times \frac{80}{63}=50\times 80=4000$ Kanchan's cost price is ₹4,000, so the correct option is (a).

Simple interest - finding principal Formula: SI $=\frac{P\times R\times T}{100}$ $177=\frac{P\times 2\times 5}{100}=\frac{10P}{100}=\frac{P}{10}$ $P=177\times 10=1770$ The principal is ₹1,770, so the correct option is (d).

Comparing two numbers by percentage Take the third number as 100, because both reductions are measured from it. First number =100-35=65 Second number =100-45=55 The question asks for the second as a percentage of the first, so the first number is the base. Required percentage $=\frac{55}{65}\times 100=\frac{1100}{13}$ $\frac{1100}{13}=84$ with remainder 8, i.e. $84\frac{8}{13}\%$ So the correct option is (d).

LCM by prime factorisation Factorise each number and keep the highest power of every prime that occurs anywhere. $40=2^3\times 5$ $20=2^2\times 5$ $195=3\times 5\times 13$ $160=2^5\times 5$ The highest powers are $2^5$, 3, 5 and 13. LCM $=32\times 3\times 5\times 13=480\times 13=6240$ So the correct option is (c).

Pipes and cisterns Take the tank as the LCM of 36, 48 and 18, that is 144 units, so every rate comes out a whole number. First pipe $=\frac{144}{36}=+4$ units/min, second $=\frac{144}{48}=+3$ units/min, third $=\frac{144}{18}=-8$ units/min. For the first 16 minutes only the two filling pipes run, so the water collected is $\left(4+3\right)\times 16=112$ units. Once all three are open the net rate is 4+3-8=-1 unit per minute, so the tank now loses 1 unit every minute. Time to remove those 112 units $=\frac{112}{1}=112$ minutes. So the correct option is (b).

Compound interest - equal amounts Nitin's money compounds for 16 years and Pravin's for 17 years, and the two final amounts are equal. $N{\left(1+\frac{10}{100}\right)}^{16}=P{\left(1+\frac{10}{100}\right)}^{17}$ Cancel ${\left(\frac{11}{10}\right)}^{16}$ from both sides: $N=P\times \frac{11}{10}$ So N:P=11:10 - the son whose money gets one year less has to start with more. Total parts =11+10=21, and these 21 parts are worth ₹1,617, so one part $=\frac{1617}{21}=77$. Nitin's share $=11\times 77=847$ Veer gave Nitin ₹847, so the correct option is (a).

Speed and time in inverse ratio The distance is fixed, so speed and time are inversely proportional: cut the speed and the time goes up in the reciprocal ratio. Reduced speed : normal speed =2:3, so reduced time : normal time =3:2. The delay is the difference 3-2=1 part, and that one part equals 47 minutes. Normal time =2 parts $=2\times 47=94$ minutes So the correct option is (b).

Volume of a cylinder Formula: volume of a cylinder $=\pi r^{2}h$, the area of the circular base times the height. $=\frac{22}{7}\times 7\times 7\times 10.2$ One factor of 7 cancels with the 7 in the denominator: $=22\times 7\times 10.2$ $=154\times 10.2=1570.8$ The volume is 1570.8 cm³, so the correct option is (a).

Price, consumption and expenditure Expenditure = price $\times$ consumption, so consumption = expenditure $÷$ price. Price goes $100\to 180$ and expenditure is allowed to go $100\to 117$. New consumption $=\frac{117}{180}\times 100=65$ Fall in consumption =100-65=35 Reduction $=\frac{35}{100}\times 100=35\%$ So the correct option is (c).

Mean proportional If x is the mean proportional between a and b then a:x=x:b, so $x^2=ab$ and $x=\sqrt{ab}$. $x=\sqrt{5.76\times 0.88}=\sqrt{5.0688}$ Since $\sqrt{5.76}=2.4$, this is $x=2.4\times \sqrt{0.88}=2.4\times 0.9381$ $x=2.2514\dots \approx 2.25$ So the correct option is (a).

Discount slabs First decide which slab the purchase falls in: ₹6,000 is above ₹5,000, so the 20% rate applies and the 10% slab is irrelevant. Saving $=\frac{20}{100}\times 6000=1200$ The customer saves ₹1,200, so the correct option is (c).

Direct proportion In direct proportion the rate per second is the same for both advertisements, so $\frac{R_1}{T_1}=\frac{R_2}{T_2}$. Rate per second $=\frac{125000}{20}=6250$ $R_2=6250\times 25=156250$ The rate is ₹1,56,250, so the correct option is (c).

Average of a small data set Formula: average = sum of the values $÷$ number of values. Sum =15000+25000+30000+12000+18000=100000 Average $=\frac{100000}{5}=20000$ The average salary is ₹20,000, so the correct option is (b).

Compound interest - equal maturity Amit is 11, so his money grows for 23-11=12 years; Gopal is 12, so his grows for 23-12=11 years. The two maturity amounts are equal, with $1+\frac{10}{100}=\frac{11}{10}$ each year. $A{\left(\frac{11}{10}\right)}^{12}=G{\left(\frac{11}{10}\right)}^{11}$ Cancel ${\left(\frac{11}{10}\right)}^{11}$: $A\times \frac{11}{10}=G$, so A:G=10:11. Total parts =10+11=21, and these 21 parts are worth ₹23,100, so one part $=\frac{23100}{21}=1100$. Gopal's share $=11\times 1100=12100$ Rodney gave Gopal ₹12,100, so the correct option is (c).