13 February 2025 · Shift 2

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$. Both sums run for the same T=6 years at the same rate R, so the whole gap of ₹600 comes from the gap in principal. Difference of principals =9400-5400=4000. So $\frac{4000\times R\times 6}{100}=600$. $240R=600\Rightarrow R=2.5$. Hence the rate of interest is $2.5\%$ per annum, so the correct option is (d).

LCM application They stand together again after a time that is a whole number of rounds for each of them, i.e. after the LCM of 6, 8, 10 and 14 minutes. $6=2\times 3$, $8=2^3$, $10=2\times 5$, $14=2\times 7$. Taking the highest power of each prime, LCM $=2^3\times 3\times 5\times 7=840$ minutes. 840 minutes $=\frac{840}{60}=14$ hours. Hence they meet again at the starting position after 14 hours, so the correct option is (d).

HCF with remainders If the divisor leaves remainder 15 in 1163, it divides 1163-15=1148 exactly. In the same way it divides 2214-82=2132 exactly. So the required number is the HCF of 1148 and 2132. $2132=1148\times 1+984$; $1148=984\times 1+164$; $984=164\times 6+0$. So HCF =164, and 164 is larger than both remainders, as a divisor must be. Hence the greatest such number is 164, so the correct option is (d).

Percentage increase Increase in salary =57000-42000=15000. Percentage increase $=\frac{\mathrm{increase}}{\mathrm{original}\mathrm{salary}}\times 100$. $=\frac{15000}{42000}\times 100=\frac{250}{7}=35.714\dots$ Rounded off to one decimal place this is $35.7\%$. The base is always the original salary ₹42,000; dividing by the new ₹57,000 gives the trap value $26.3\%$. Hence the correct option is (a).

Surface area of hemisphere A solid hemisphere shows two surfaces: the curved dome and the flat circular face. Total surface area $=2\pi r^2+\pi r^2=3\pi r^2$. r=0.07 m =7 cm, and the options are in cm², so work in cm. TSA $=3\times \frac{22}{7}\times 7^2=3\times 22\times 7=462$. Hence the total surface area is 462 cm², so the correct option is (d).

LCM of numbers The smallest number divisible by all of them is their LCM. 5=5, $10=2\times 5$, $49=7^2$, $28=2^2\times 7$. Take the highest power of every prime that appears: $2^2\times 5\times 7^2$. LCM $=4\times 5\times 49=980$. Checking the traps: 1040 and 900 are not divisible by 49, and 1006 is not divisible by 5. Hence the smallest such natural number is 980, so the correct option is (a).

Profit percentage Let the selling price of one chair be S and its cost price be C. Gain on 150 chairs = selling price of 50 chairs, so 150S-150C=50S. $100S=150C\Rightarrow \frac{S}{C}=\frac{3}{2}$; take S=3 and C=2. Profit $\%=\frac{S-C}{C}\times 100=\frac{3-2}{2}\times 100=50\%$. The gain is always measured on the cost price, so the answer is not $\frac{1}{3}\times 100$. Hence his profit percentage is $50\%$, so the correct option is (b).

Compound interest ratio Let Shan get ₹S and Piyush get ₹P, so S+P=1612. Their amounts are equal: $S{\left(1+\frac{8}{100}\right)}^{19}=P{\left(1+\frac{8}{100}\right)}^{20}$. Dividing both sides by ${\left(1+\frac{8}{100}\right)}^{19}$ gives $S=P\left(1+\frac{8}{100}\right)$. $\frac{S}{P}=\frac{27}{25}$, so the money is divided in the ratio 27:25, i.e. 27+25=52 parts. One part $=\frac{1612}{52}=31$. Shan's share $=27\times 31=837$. Hence Rishi gave ₹837 to Shan, so the correct option is (b).

Work and wages Take the total work as the LCM of 5, 8 and 6, i.e. 120 units. One day's work: A $=\frac{120}{5}=24$, B $=\frac{120}{8}=15$, C $=\frac{120}{6}=20$ units. Wages are shared in the ratio of work done, i.e. 24:15:20, a total of 59 parts. One part $=\frac{22774}{59}=386$. Difference of B's and C's shares $=\left(20-15\right)\times 386=5\times 386=1930$. Hence the difference is ₹1,930, so the correct option is (a).

Selling price from profit Formula: $SP=CP\times \left(1+\frac{\mathrm{profit}\%}{100}\right)$. $SP=500\times \left(1+\frac{24}{100}\right)=500\times \frac{124}{100}$. $SP=5\times 124=620$. Hence he should sell the article for ₹620, so the correct option is (d).

HCF and LCM HCF =7, so LCM $=126\times 7=882$. Let the numbers be 7a and 7b with a and b coprime; then LCM $=7ab=882\Rightarrow ab=126$. Their difference is $5\times 7=35$, so $7\left(a-b\right)=35\Rightarrow a-b=5$. ab=126 with a-b=5 gives a=14 and b=9, which are coprime as required. Product $=7a\times 7b=49\times 126=6174$. The same value follows from HCF $\times$ LCM $=7\times 882=6174$. Hence the product of the two numbers is 6174, so the correct option is (b).

Average of weights Sum of the three weights $=3\times 46=138$ kg. Anwar + Amit $=2\times 39=78$, so Gopal =138-78=60 kg. Amit + Gopal $=2\times 44=88$, so Anwar =138-88=50 kg. Amit =138-60-50=28 kg. Hence the weight of Amit is 28 kg, so the correct option is (d).

Successive percentage change Take the original price as 100. A fall of $x\%$ multiplies the price by $\left(1-\frac{x}{100}\right)$ and a rise by $\left(1+\frac{x}{100}\right)$. Final price $=100\times 0.6\times 0.7\times 0.9\times 1.4=52.92$. Change =52.92-100=-47.08 on a base of 100. Percentages of different months cannot be added, which is why $-40-30-10+40=-40\%$ is wrong. Hence the price decreases by $47.08\%$, so the correct option is (a).

SI and CI difference For 2 years SI =42, so the interest of one year $=\frac{42}{2}=21$. For 2 years, CI - SI is exactly one year's interest on the first year's interest. 63-42=21, so $21\times \frac{R}{100}=21\Rightarrow R=100$. Now $SI=\frac{P\times R\times T}{100}$ gives $\frac{P\times 100\times 2}{100}=42\Rightarrow 2P=42$. P=21. Check: at $100\%$ the amount grows $21\to 42\to 84$, so CI =84-21=63, as given. Hence the principal is ₹21, so the correct option is (a).

Alligation of prices Rule of alligation: $\frac{\mathrm{cheaper}\mathrm{quantity}}{\mathrm{dearer}\mathrm{quantity}}=\frac{\mathrm{dearer}\mathrm{price}-\mathrm{mean}\mathrm{price}}{\mathrm{mean}\mathrm{price}-\mathrm{cheaper}\mathrm{price}}$. $=\frac{285-195}{195-165}=\frac{90}{30}=\frac{3}{1}$. The mean 195 lies much nearer to 165, so the cheaper variety must be the larger share, which 3:1 satisfies. Hence the two varieties must be mixed in the ratio 3:1, so the correct option is (b).

Relative speed The man runs in the same direction as the train, so use the relative speed =88-16=72 km/h. 72 km/h $=72\times \frac{5}{18}=20$ m/s. A train passes a man when it covers its own length, 240 m, at this relative speed. Time $=\frac{\mathrm{length}}{\mathrm{relative}\mathrm{speed}}=\frac{240}{20}=12$ seconds. Hence the train passes him in 12 seconds, so the correct option is (b).

Fourth proportional If x is the fourth proportional of a, b, c then a:b=c:x. $4:12=20:x\Rightarrow 4x=12\times 20$. $x=\frac{240}{4}=60$. Hence the fourth proportional is 60, so the correct option is (b).

Marked price and discount A discount of $20\%$ means the pair was paid for at $80\%$ of the marked price of the pair. $\frac{80}{100}\times M{P}_{\mathrm{pair}}=1820$. $M{P}_{\mathrm{pair}}=\frac{1820\times 100}{80}=2275$. Both articles carry the same marked price, so the MP of one $=\frac{2275}{2}=1137.50$. Hence the marked price of an article is ₹1,137.50, so the correct option is (a).

Boat and stream Let the speed of the boat in still water be x km/h; then downstream =x+9 and upstream =x-9. $\frac{28}{x+9}+\frac{20}{x-9}=8$. 28(x-9)+20(x+9)=8(x+9)(x-9). $48x-72=8x^2-648$. $8x^2-48x-576=0\Rightarrow x^2-6x-72=0$. (x-12)(x+6)=0, and a speed cannot be negative, so x=12. Hence the speed of the boat in still water is 12 km/h, so the correct option is (c).

Time and work (men-days) The job is fixed, so men $\times$ days stays constant: $M_1{D}_1=M_2{D}_2$. $80\times 100=M_2\times 25$. $M_2=\frac{8000}{25}=320$ men are needed to finish it in 25 days. The question asks for the extra men: 320-80=240. Hence 240 more men should be employed, so the correct option is (d).