20 February 2025 · Shift 3

Simple and compound interest For 2 years the only extra amount in compound interest is the interest earned on the first year's interest, so $CI-SI=SI\times \frac{R}{200}$. Here CI-SI=80-40=₹40, so $40=40\times \frac{R}{200}$. $\Rightarrow R=200\%$ per annum. Principal $=\frac{SI\times 100}{R\times T}=\frac{40\times 100}{200\times 2}=₹10$. Check: at $200\%$, ₹10 gives SI=₹40 and amount $=10\times 3\times 3=₹90$, i.e. CI=₹80 - both match. Hence the correct option is (b).

Average speed The same distance is covered twice, so the average speed is the harmonic mean, not the ordinary mean: $\mathrm{Average}\mathrm{speed}=\frac{2xy}{x+y}$. Substituting x=22 and y=17: $=\frac{2\times 22\times 17}{22+17}$ $=\frac{748}{39}$ $=19\frac{7}{39}$ km/hr. The tempting $\frac{22+17}{2}=19.5$ km/hr is wrong because more time is spent at the slower speed, so the true average must be below 19.5. Hence the correct option is (a).

Time and work P does $\frac{1}{3}$ of the work in 4 days, so P alone needs $3\times 4=12$ days. Q does $\frac{1}{4}$ of the work in 6 days, so Q alone needs $4\times 6=24$ days. Take the whole work $=\mathrm{LCM}\left(12,24\right)=24$ units; then P does 2 units/day and Q does 1 unit/day. In 6 days P and Q together do $6\times \left(2+1\right)=18$ units. Work left for R =24-18=6 units, done in 2 days, so R does $\frac{6}{2}=3$ units/day. R alone $=\frac{24}{3}=8$ days. Hence the correct option is (d).

HCF of numbers A scale measures a side exactly only if its length divides that side, so the largest such scale is the HCF of the four sides. $95=5\times 19$, $76=2^2\times 19$, $57=3\times 19$, $152=2^3\times 19$ The only factor common to all four is 19, so $\mathrm{HCF}=19$ cm. 27 and 11 divide none of the sides and 31 divides none either, so only 19 works. Hence the correct option is (c).

Speed, time and distance The distance is fixed, so time is inversely proportional to speed: $S\propto \frac{1}{T}$. Speeds are in the ratio normal : reduced $=1:\frac{2}{3}=3:2$, so the times are in the ratio 2:3. The delay is 3-2=1 unit of time, and this equals 30 minutes. Normal time =2 units $=2\times 30=60$ minutes. Hence the correct option is (c).

Surface area of a sphere Formula: surface area of a sphere $=4\pi r^2$ (the curved surface is the whole surface here). Substituting r=5.4 cm: $=4\times \pi \times {\left(5.4\right)}^2$ $=4\pi \times 29.16$ $=116.64\pi {\mathrm{cm}}^2$ So the surface area is $116.64\pi {\mathrm{cm}}^2$, and the correct option is (d).

Profit and selling price Cost price of 100 chickoos =₹1210 (the rate is given per hundred). Selling price of 100 chickoos $=CP+\mathrm{Profit}=1210+940=₹2150$. Selling price of one chickoo $=\frac{2150}{100}=₹21.50$. Selling price per dozen $=21.50\times 12=₹258$. Hence the correct option is (b).

BODMAS with integers By BODMAS, division and multiplication are done before the additions and subtractions. Division: $\left(-60\right)÷\left(-15\right)=+4$ (a negative divided by a negative is positive). Multiplication: $\left(-3\right)\times 8=-24$. So the expression =(-9)-(+4)+(-24) =-9-4-24=-37 The common slip is to keep -(-60) together and get +4 added instead of subtracted, which gives -29. Hence the correct option is (d).

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$; here R and T=5 are the same for both, so the difference of interests comes only from the difference of principals. $\frac{\left(9400-5400\right)\times R\times 5}{100}=480$ $\frac{4000\times 5\times R}{100}=480$ 200R=480 $R=2.4\%$ per annum. Hence the correct option is (d).

Effective discount In a buy-2-get-1-free offer the customer pays for 2 pens but carries home 3, so the discount must be compared with the price of all 3 pens. Marked price of 3 pens $=3\times 10=₹30$; amount paid $=2\times 10=₹20$; discount =₹10. Discount $\%=\frac{\mathrm{discount}}{\mathrm{marked}\mathrm{price}}\times 100=\frac{10}{30}\times 100$ $=\frac{100}{3}=33\frac{1}{3}\%$ Dividing by ₹20 instead of ₹30 gives $50\%$ - the discount is always taken on the marked price of everything received. Hence the correct option is (d).

BODMAS simplification Work from the innermost bracket outwards: (6-5)=1. Inside the curly bracket, multiply before subtracting: $18-6\times 1\times 3=18-18=0$. $8\times 0=0$, and then $0÷36=0$. Square bracket: [3+0]=3. $36-10\times 3=36-30=6$ Hence the correct option is (d).

Difference of CI and SI For 2 years, $CI-SI=SI\times \frac{R}{200}$ - the extra ₹ is only the interest on the first year's interest. CI-SI=41-40=₹1, so $1=40\times \frac{R}{200}$. $\Rightarrow R=5\%$ per annum. Principal $=\frac{SI\times 100}{R\times T}=\frac{40\times 100}{5\times 2}=₹400$. Check: ₹400 at $5\%$ gives SI=₹40 and $CI=400\left({\left(1.05\right)}^2-1\right)=₹41$. Hence the correct option is (b).

Successive percentage change The second change acts on the increased price, not on the original one, so multiply the two factors. Let the original cost price be x: $x\times \frac{115}{100}\times \frac{85}{100}=19550$ $x\times \frac{23}{20}\times \frac{17}{20}=19550$ $x=\frac{19550\times 20\times 20}{23\times 17}=\frac{19550\times 400}{391}$ $x=50\times 400=₹20,000$ Hence the correct option is (a).

Loss percentage Loss is always reckoned on the cost price, so $SP=CP\times \frac{100-\mathrm{loss}\%}{100}$. $SP=2200\times \frac{100-35}{100}=2200\times \frac{65}{100}$ $=22\times 65=₹1430$ Hence the correct option is (b).

Mean proportional If b is the mean proportional of a and c, then a:b=b:c, so $b^2=ac$ and $b=\sqrt{ac}$. $b=\sqrt{0.5\times 72}$ $=\sqrt{\frac{1}{2}\times 72}=\sqrt{36}$ =6 Hence the correct option is (d).

Average and a new member Total age of the 52 students $=52\times 12=624$ years. With the teacher the group has 53 members: total age $=53\times 13=689$ years. Teacher's age =689-624=65 years. Short trick: the teacher's age = new average + (rise in average $\times$ number of students) $=13+1\times 52=65$ years. Hence the correct option is (d).

LCM of numbers Write each number in prime factors and take the highest power of every prime. $44=2^2\times 11$, $32=2^5$, $176=2^4\times 11$, $352=2^5\times 11$ $\mathrm{LCM}=2^5\times 11=32\times 11=352$ Note that 352 is itself divisible by 44, 32 and 176, so the largest number is the LCM here. Hence the correct option is (c).

Third proportional If c is the third proportional of a and b, then a:b=b:c, so $c=\frac{b^2}{a}$. Third proportional of 32 and 48 $=\frac{{48}^2}{32}=\frac{2304}{32}=72$ Third proportional of 4 and 18 $=\frac{{18}^2}{4}=\frac{324}{4}=81$ Required ratio =72:81=8:9 (dividing both by 9). Hence the correct option is (c).

Percentage in population Number of females =357000-194000=163000. Let $x\%$ of the females be literate. Literate males + literate females = literate population: $\frac{47}{100}\times 194000+\frac{x}{100}\times 163000=\frac{47}{100}\times 357000$ $\frac{x}{100}\times 163000=\frac{47}{100}\times \left(357000-194000\right)=\frac{47}{100}\times 163000$ $x=47\%$ In general, when the whole and one part have the same percentage, the remaining part must carry that very percentage too. Hence the correct option is (b).

Meeting on a circular track They run in the same direction, so what matters is the relative speed =45-36=9 km/hr. 9 km/hr $=9\times \frac{5}{18}=2.5$ m/s. They meet again only when the faster runner has gained one full lap on the other, i.e. 500 m. Time $=\frac{\mathrm{lap}}{\mathrm{relative}\mathrm{speed}}=\frac{500}{2.5}=200$ seconds. Hence the correct option is (c).