Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for Rs 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in Rs received by him is (in numerical value)
Since money paid is in proportion to the work done Let work = 24000 units (1 unit = 1 Re) Anil’s efficiency = 24000/12 = 2000 units/day Barun’s efficiency = 24000/16 = 1500 units/day Together, Anil + Barun + Chandu = 24000/6 = 4000 units/day Chandu’s efficiency = 4000 – 2000 – 1500 = 500 units/day Since Chandu worked for 6 days, money paid = 500 × 6 = Rs 3000
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is (in numerical value)
Sum (25 students) = 25 × 50 = 1250 Let the score of each topper = x, sum (5 toppers) = 5x Remaining students = 20 To maximize the score of the topper, we have to minimize the remaining score with 30 being least and all distinct integer Sum (20 students minimum) = 30 + 31 + 32 + … 20 values = 20/2 (2 × 30 + 19 × 1) = 790 So, Sum (5 toppers) = 5x = 1250 – 790 = 460 Hence, the score of each topper (maximum) = 460/5 = 92
A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was Rs 50 less than that of a large shirt. She paid a total of Rs 5000 for the large shirts, and a total of Rs 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in Rs, is
Let the number of large size shirt = x and small size shirt = 64 – x Let the price of the large shirt = P, then price of small shirt = P – 50 Given, P × x = 5000 and (P – 50)(64 – x) = 1800 64P – Px – 3200 + 50x = 1800 64P + 50x = 10000 32P + 25 (5000/P) = 5000 32P2 – 5000P + 125000 = 0 4P2 – 625P + 15625 = 0 4P2 – 500P – 125P + 15625 = 0 (4P – 125)(P – 125) = 0 P = 125 (as P = 125/4 making P – 50 negative, so rejected) Hence the price of large shirt and small shirt together = P + (P – 50) = 200
The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is
A tea shop offers tea in cups of three different sizes. The product of the prices, in Rs, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by Rs 6, keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in Rs, is (in numerical value)
If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is
Let the weight of the initial alloy = x kg and percentage of silver in the alloy = p% Given, p% of x + 100% of 3 = 90% of (x + 3) px + 300 = 90x + 270, 90x – px = 30 Also, p% of x + 90% of 2 = 84% of (x + 2) px + 180 = 84x + 168, 84x – px = 12 Subtracting and solving, 6x = 18, x = 3 kg
One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be
Let the price of smallest cup = 2x, price of medium cup = 5x And let the price of largest cup = p Given, 2x × 5x × p = 800, p = 80/x2 Also, (2x + 6) × (5x + 6) × p = 3200 => (10x2 + 42x + 36) × p = 3200 => 10x2 + 42x + 36 = 40x2 => 5x2 – 7x – 6 = 0 => (5x + 3) (x – 2) = 0 => x = 2 and p = 20 Required sum = 2x + 5x + p = 34
In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be
Matches played = 40, Won = 30% of 40 = 12 Let remaining matches = x, Win among them = 60% of x = 0.6 Overall win, 12 + 0.6x = 50% of (40 + x) Solving, x = 80 = remaining matches Required value = 12 + 90% of 80 = 84
The cost of fencing a rectangular plot is Rs 200 per ft along one side, and Rs 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in Rs, is
Let L and B be the length and breadth of rectangular plot One side cost = Rs 200 be on one of the lengths and other three sides cost = Rs 100 Total cost = 200L + 100L + 100B + 100B = 300L + 200B The cost to be lowest possible, 300L = 200B = k Area, L × B = 60000, k/300 × k/200 = 60000, k = 60000 Hence, L = 200 and B = 300 Required lowest cost = 300 × 200 + 200 × 300 = Rs 120000
One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is
Let R and G be the efficiencies respectively Given, R × 8 + G × 6 = W (total work) × 5 Also, R × 7.5 + G × 7.5 = W × 4 Subtracting, 40R – 30R = W, R = W/10 Hence, Rahul alone would have taken 10 hours
Bank A offers 6% interest rate per annum compounded half yearly. Bank B and Bank Coffer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests Rs 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in Rs, to Rupa is
Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is (in numerical value)
Let the length of the track = x Let the speed of Mira and Amal be m and a respectively If Amal completes 3 more round than Mira in 45 min walking in same direction, a – m = 3x/45 = x/15 Also, when they walk in opposite direction, a + m = x/3 Solving, a = x/5 and m = 2x/15 = x/7.5 So, Mira walks one round in 7.5 mins Hence number of rounds Mira walks in one hours = 60/7.5 = 8
In a triangle ABC, angle BCA = 50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then angle FDE, in degrees, is equal to
Given, angle BCA = 50°
AD = DE and BD = DF
Let angle DAE = angle DEA = x
So, angle ADE = 180° – 2x
Also let angle BDF = angle BFD = y°
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So, angle BDF = 180° – 2y
Also, angle A + B + C = 180°
So, x + y = 130°
Again, angle ADE + angle BDF + angle FDE = 180°
180° – 2x + 180° – 2y + angle FDE = 180°
Angle FDE = 260° – 180° = 80°
A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is (in numerical value)
Following cases are possible (2, 3, 1, 1) can be arranged in 4!/2! = 12 cases (2, 3, 1, 2) can be arranged in 4!/2! = 12 cases (2, 3, 1, 3) can be arranged in 4!/2! = 12 cases (2, 3, 2, 2) can be arranged in 4!/3! = 4 cases (2, 3, 2, 3) can be arranged in 4!/2!2! = 6 cases (2, 3, 3, 3) can be arranged in 4!/3! = 4 cases Total = 50 cases
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