19 February 2025 · Shift 3

Successive percentage change Successive percentage changes are multiplied, never added, so take the original price as ₹100. A fall of 20% keeps $\frac{80}{100}$, a fall of 35% keeps $\frac{65}{100}$, a fall of 20% keeps $\frac{80}{100}$ and a rise of 50% makes it $\frac{150}{100}$ times. Final price $=100\times \frac{80}{100}\times \frac{65}{100}\times \frac{80}{100}\times \frac{150}{100}=62.4$ Change =100-62.4=37.6, and it is a fall because the final price is below ₹100. Percentage decrease $=\frac{37.6}{100}\times 100=37.6$, i.e. the price decreases by 37.6%. Therefore the correct option is (a).

Trains, relative speed To cross another train completely, the faster train must cover both lengths. Distance =370+330=700 m The trains move in the same direction, so the relative speed is the difference of the speeds. Relative speed =150-90=60 km/h $=60\times \frac{5}{18}=\frac{50}{3}$ m/s Time $=\frac{700}{\frac{50}{3}}=\frac{700\times 3}{50}=42$ seconds In minutes, $\frac{42}{60}=0.7$ Therefore the correct option is (d).

BODMAS with integers By BODMAS, division and multiplication are done before subtraction and addition. $\left(-60\right)÷\left(-15\right)=+4$ (a negative divided by a negative is positive) $\left(-4\right)\times 9=-36$ So the expression becomes (-9)-(+4)+(-36) =-9-4-36=-49 Therefore the correct option is (a).

HCF with remainders If a divisor leaves remainder 40 in 2340, it divides 2340-40=2300 exactly. Likewise it divides 1551-33=1518 exactly. The greatest such divisor is the HCF of 2300 and 1518. $2300=2^2\times 5^2\times 23$ and $1518=2\times 3\times 11\times 23$ HCF $=2\times 23=46$ 46 is bigger than both remainders 40 and 33, so it is a valid divisor. Therefore the correct option is (d).

Curved surface area of a cone Lateral (curved) surface area of a cone $=\pi rl$, where r is the radius and l the slant height. Both must be in the same unit, so convert the slant height: 0.21 m $=0.21\times 100=21$ cm. Area $=\frac{22}{7}\times 3.5\times 21$ $=22\times 0.5\times 21=231$ So the lateral surface area is 231 cm${}^2$. Therefore the correct option is (a).

Cost price from gain percent Gain per cent is always reckoned on the cost price, not on the selling price. A gain of 12.5%, i.e. $\frac{1}{8}$, makes SP = CP $+\frac{1}{8}$ CP $=\frac{9}{8}$ CP. $216=\frac{9}{8}\times$ CP CP $=216\times \frac{8}{9}=192$ So the cost price was ₹192. Therefore the correct option is (d).

Simple interest, difference Simple interest $=\frac{P\times R\times T}{100}$, and here R and T are the same for both men. So the extra interest comes only from the extra principal, 9400-5400=4000. Difference $=\frac{4000\times R\times 5}{100}=200R$ Given, 200R=600 $R=\frac{600}{200}=3$ So the rate of interest is 3% per annum. Therefore the correct option is (a).

BODMAS with integers By BODMAS, do the division and the multiplication first. $\left(-60\right)÷\left(-15\right)=+4$ $\left(-4\right)\times 7=-28$ The expression becomes (-9)-(+4)+(-28) =-9-4-28=-41 Therefore the correct option is (c).

Combined average Averages of unequal groups cannot be averaged directly; convert each to a total first. Total of the first 12 $=12\times 33=396$ Total of the next 8 $=8\times 45=360$ Class total =396+360=756 for 12+8=20 students Average $=\frac{756}{20}=37.8$ The combined average must lie between 33 and 45, which rules out 45.6 and 46.5 straight away. Therefore the correct option is (c).

Average weight, pair sums Turn every average into a sum, because sums can be added and subtracted. Mohan + Mohammad + Ashish $=3\times 45=135$ Mohan + Mohammad $=2\times 43=86$ Mohammad + Ashish $=2\times 46=92$ Adding the last two counts Mohammad twice: 86+92=178=135+ Mohammad Mohammad =178-135=43 So Mohammad weighs 43 kg, and the correct option is (c).

CI and SI for two years For the same principal, rate and time, CI and SI are in a fixed ratio, so the principal need not be found separately. For 2 years, CI $=P\left({\left(1+\frac{4}{100}\right)}^2-1\right)=P\left(\frac{676}{625}-1\right)=\frac{51P}{625}$ For 2 years, SI $=\frac{P\times 4\times 2}{100}=\frac{8P}{100}=\frac{50P}{625}$ So CI : SI =51:50 (the extra 1 part is the interest earned on the first year's interest). SI $=280.5\times \frac{50}{51}$ $=5.5\times 50=275$ So the simple interest is ₹275, and the correct option is (d).

Profit, price per dozen ₹1250 per hundred means 400 pears are four hundreds. Cost price $=1250\times 4=5000$, i.e. ₹5,000 Selling price = cost price + profit =5000+900=5900, i.e. ₹5,900 Selling price of one pear $=\frac{5900}{400}=14.75$ Selling price per dozen $=14.75\times 12=177$ So one dozen sells for ₹177, and the correct option is (b).

Average speed, equal distance The two legs cover the same distance, so the average speed is the harmonic mean of the speeds. Average speed $=\frac{2xy}{x+y}$, where x=20 and y=30 $=\frac{2\times 20\times 30}{20+30}$ $=\frac{1200}{50}=24$ It is not the plain average $\frac{20+30}{2}=25$, because more time is spent on the slower leg. Therefore the correct option is (b).

Discount percentage Discount = marked price - selling price =3500-2800=700 Discount per cent is taken on the marked price, never on the selling price. Discount % $=\frac{700}{3500}\times 100$ $=\frac{1}{5}\times 100=20$ So the discount is 20%, and the correct option is (a).

Time and work, units method Take the total work as the LCM of 20, 30 and 36, which is 180 units. One day's work: $X=\frac{180}{20}=9$, $Y=\frac{180}{30}=6$, $Z=\frac{180}{36}=5$ units. Y and Z together do 6+5=11 units a day, so in 5 days they do $11\times 5=55$ units. Remaining work =180-55=125 units X needs $\frac{125}{9}=13\frac{8}{9}$ days A part of a day still counts as a whole working day, so X takes 14 complete days. Therefore the correct option is (a).

Ratio and difference Let the numbers be 16x and 23x, so that their ratio stays 16:23. Difference =23x-16x=7x 7x=224 $x=\frac{224}{7}=32$ Smaller number $=16x=16\times 32=512$ Therefore the correct option is (d).

Inverse variation 'x varies inversely as the square of y' means $x=\frac{k}{y^2}$, where k is a constant. Put the given pair x=6, y=5 to find k: $6=\frac{k}{5^2}=\frac{k}{25}$ $k=6\times 25=150$ Now put y=4: $x=\frac{150}{4^2}=\frac{150}{16}$ =9.375 Therefore the correct option is (c).

BODMAS with brackets Open the innermost bracket first: 9-2=7. Inside the curly bracket, multiplication comes before addition: $\frac{19}{7}\times 7=19$. Also $\frac{57}{3}=19$, so the curly bracket =19+19=38. Next the round bracket: $49÷7=7$. Value $=7\times 38=266$ Therefore the correct option is (b).

Price rise, consumption cut Expenditure = price $\times$ consumption, and the expenditure has to stay the same. So if the price becomes $\frac{122.5}{100}$ times, the consumption must become $\frac{100}{122.5}$ times. Reduction in consumption $=1-\frac{100}{122.5}=\frac{22.5}{122.5}$ Reduction % $=\frac{22.5}{122.5}\times 100=\frac{2250}{122.5}$ $=18.367\dots$ Correct to two decimal places, she must cut her consumption by 18.37%. Therefore the correct option is (b).

Compound interest, share division At 10% compounded annually, money is multiplied by $\frac{11}{10}$ every year. Let Piyush get P and Manoj get M; their amounts after 18 and 19 years are equal. $P{\left(\frac{11}{10}\right)}^{18}=M{\left(\frac{11}{10}\right)}^{19}$ Cancelling ${\left(\frac{11}{10}\right)}^{18}$ from both sides, $P=M\times \frac{11}{10}$ So P:M=11:10 - the one whose money grows for one year longer needs one factor less. Total parts =11+10=21, and 21 parts =1218, so 1 part $=\frac{1218}{21}=58$ Manoj's share $=10\times 58=580$ So Shan gave Manoj ₹580, and the correct option is (d).