6 February 2025 · Shift 1

Successive discounts Take the marked price as ₹100 and apply the three discounts one after the other. Price left $=100\times \left(1-\frac{30}{100}\right)\times \left(1-\frac{25}{100}\right)\times \left(1-\frac{40}{100}\right)$ $=100\times 0.70\times 0.75\times 0.60$ $=100\times 0.315=31.5$ Single discount =100-31.5=68.5 Hence the equivalent single discount is 68.5% and the correct option is (c).

Percentage - price and consumption Take the original price as ₹100 per litre and the original consumption as 1 litre, so the original expenditure is ₹100. New price $=100\times 1.50=150$ New expenditure $=100\times 1.14=114$ New consumption $=\frac{114}{150}=0.76$ litre Reduction in consumption $=\frac{1-0.76}{1}\times 100=24$ Hence the consumption must be reduced by 24% and the correct option is (d).

Time and work Ravi does $\frac{1}{4}$ of the work in 15 days, so the whole work takes him $15\times 4=60$ days. Chavi takes $18\times 3=54$ days and Dhani takes $24\times 2=48$ days for the whole work. Let the total work be the LCM of 60, 54 and 48, that is 2160 units. One day's work: Ravi $=\frac{2160}{60}=36$, Chavi $=\frac{2160}{54}=40$, Dhani $=\frac{2160}{48}=45$ Together in one day =36+40+45=121 units Three-fourth of the work $=\frac{3}{4}\times 2160=1620$ units Required time $=\frac{1620}{121}=13\frac{47}{121}$ days Hence the correct option is (a).

HCF with remainders If a number leaves remainder 7 on dividing 92, it divides 92-7=85 exactly. In the same way it divides 215-11=204 exactly. Required number = HCF of 85 and 204. $85=5\times 17$ and $204=2^2\times 3\times 17$, so the common factor is 17. 17 is greater than both remainders 7 and 11, so it is a valid divisor. Hence the required number is 17 and the correct option is (a).

HCF with remainders The number divides 406-1=405 exactly and also 1388-3=1385 exactly. $405=3^4\times 5$ and $1385=5\times 277$ The only common factor is 5, so the HCF =5. 5 is greater than both remainders 1 and 3, so it is a valid divisor. Hence the required number is 5 and the correct option is (c).

Alligation on averages Apply alligation to the three average speeds, with 35 km/hr as the mean. Yellow : green =(35-30):(55-35) =5:20=1:4 Total parts =1+4=5, and these 5 parts stand for 100 bikes. 1 part $=\frac{100}{5}=20$ bikes Green bikes $=4\times 20=80$ Hence the correct option is (b).

Speed, time and distance The distance is fixed, so time is inversely proportional to speed. Speed becomes $\frac{2}{3}$ of normal, so the time becomes $\frac{3}{2}$ of normal. Let the normal time be t minutes. Extra time $=\frac{3}{2}t-t=\frac{t}{2}$ $\frac{t}{2}=10$ t=20 Hence the normal time is 20 minutes and the correct option is (a).

Percentage of a number $\frac{4}{100}\times x=132$ $x=132\times \frac{100}{4}$ $x=132\times 25$ x=3300 Hence the correct option is (b).

Third proportional If c is the third proportional to a and b, then a:b=b:c. So $c=\frac{b^2}{a}$ $c=\frac{112\times 112}{28}$ $c=\frac{12544}{28}=448$ Hence the correct option is (d).

Successive profit and loss Work backwards from the price Akshay paid. Gopal sold at a 25% loss, so Gopal's cost price $\times \frac{75}{100}=1890$ Gopal's cost price $=\frac{1890\times 100}{75}=2520$ This ₹2520 is Amit's selling price, earned at a 5% profit. Amit's cost price $\times \frac{105}{100}=2520$ Amit's cost price $=\frac{2520\times 100}{105}=2400$ Hence the cost price is ₹2400 and the correct option is (c).

Mean proportional The mean proportional of p and q is $\sqrt{pq}$. Required value $=\sqrt{\frac{a^2}{b^3}\times \frac{9b^2}{4a^3}}$ $=\sqrt{\frac{9a^2{b}^2}{4a^3{b}^3}}$ $=\sqrt{\frac{9}{4ab}}$ $=\frac{3}{2\sqrt{ab}}$ Hence the correct option is (b).

Compound interest ratio For compound interest, amount $=P{\left(1+\frac{R}{100}\right)}^n$. Arun's amount after 14 years equals Mahesh's amount after 15 years. $P_A{\left(\frac{11}{10}\right)}^{14}=P_M{\left(\frac{11}{10}\right)}^{15}$ $\frac{P_A}{P_M}=\frac{11}{10}$ Total parts =11+10=21 1 part $=\frac{1617}{21}=77$ Arun's share $=11\times 77=847$ Hence Ashish gave ₹847 to Arun and the correct option is (c).

BODMAS simplification By BODMAS, clear the innermost bracket first. 19-13=6, so $\frac{\left(19-13\right)}{3}=\frac{6}{3}=2$ Inside the curly bracket: $3÷3\times 2=1\times 2=2$ $23\times 16=368$ $368\times 2=736$ Hence the correct option is (b).

Compound interest, 10-month cycle $2\frac{1}{2}$ years =30 months, and interest is compounded every 10 months. Number of cycles $=\frac{30}{10}=3$ Rate per cycle $=12\times \frac{10}{12}=10$ Amount $=x{\left(1+\frac{10}{100}\right)}^3=x\times \frac{1331}{1000}$ $x\times \frac{1331}{1000}=54571$ $x=\frac{54571\times 1000}{1331}=41000$ Hence the sum is ₹41000 and the correct option is (a).

Average of weights Let the weights of Rupesh, Mandar and Ketan be R, M and K kg. $R+M+K=45\times 3=135$ $R+M=43\times 2=86$ $M+K=45\times 2=90$ Adding the last two: (R+M)+(M+K)=86+90=176 That is (R+M+K)+M=176 135+M=176 M=41 Hence Mandar weighs 41 kg and the correct option is (b).

Integers and BODMAS By BODMAS, do the division and the multiplication before the addition and subtraction. $\left(-60\right)÷\left(-15\right)=4$ $\left(-4\right)\times 6=-24$ The expression becomes (-9)-4+(-24) =-9-4-24 =-37 Hence the correct option is (b).

Trains, time and distance Let the distance be d km. The two runs differ by 10-4=6 minutes $=\frac{1}{10}$ hour. $\frac{d}{40}-\frac{d}{45}=\frac{1}{10}$ $d\left(\frac{45-40}{1800}\right)=\frac{1}{10}$ $\frac{d}{360}=\frac{1}{10}$ d=36 km Time at 40 km/hr $=\frac{36}{40}\times 60=54$ minutes, which is 10 minutes late. Correct time =54-10=44 minutes Hence the correct option is (c).

Surface area of a hemisphere The bowl is painted on the inner curved surface and on the outer curved surface, so two curved surfaces are covered. Painted area $=2\times 2\pi r^2=4\pi r^2$ $=4\times \frac{22}{7}\times 3.5\times 3.5$ =154 square cm Cost $=\frac{154}{10}\times 15$ $=15.4\times 15=231$ Hence the cost of painting is ₹231 and the correct option is (c).

Equal selling price, profit and loss When two articles are sold at the same price, one at x% profit and the other at x% loss, the result is always a loss. Loss percentage $=\frac{x^2}{100}$ $=\frac{22\times 22}{100}$ $=\frac{484}{100}=4.84$ Hence Gill has a loss of 4.84% and the correct option is (b).

Simple interest Simple interest $=\frac{P\times R\times T}{100}$ The rate and the time are the same for both, so the extra interest comes only from the extra principal. Extra principal =9400-5400=4000 $\frac{4000\times R\times 3}{100}=960$ 120R=960 R=8 Hence the rate of interest is 8% per annum and the correct option is (a).