Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is (in numerical value)
Let AD = 2x, BD = x
AE = 2y, EC = 3y
Area of ∆AED = 8
⇒ 1/2 × 2x × 2y × SinA = 8
⇒2xy Sin A = 8
⇒ xy Sin A = 4 ----(i)
Area of ∆ABC = 1/2 × 3x × 5y Sin A
= 1/2 × 15 × xy Sin A
= 1/2 × 15 × 4 = 30 Sq unit
Two trains A and B were moving in opposite DIRECTIONSs, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is
Let the speeds of trains A and B is 5x & 3x respectively and their lengths be l1 & l2 respectively.
In the first case, distance travelled will be equal to length of train B
Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by Rs. 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by Rs. 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is
Here the investment of Anil is 70% of the total investment. Now his share of profit decreased by Rs 420 when the profit decreased from 18% to 15%. So 70% of (decreased profit) = 420 ⇒ Decreased profit = 420/0.7 = Rs 600. Therefore, 3% of total investment = 600 ⇒ Total investment = Rs 20000 So, Anil’s investment is 70% of 20000 = Rs 14000. Now Chintu’s profit increased by Rs 80 when the profit increases from 15% to 17%. Therefore, 2% of Chintu’s investment = 80 ⇒ Chintu’s investment = 80/0.02 = Rs 4000 Hence, Bobby’s investment = 20000 – (14000 + 4000) = Rs 2000.
The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is
Join AP
Here in ∆ADP & ∆BPC,
AD = BP, DP = PC
∠BPC = ∠PBA = ∠ADP
⇒ ∆s ADP & BPC are congruent
Also AP is diagonal of parallelogram ABPD
∴ Ar (∆ADP) = Ar (∆ABP)
Let Ar (∆ADP) = x = Ar (∆ABP) = Ar (∆BPC)
∴ Area of trapezium ABCD = 3x
Now given that 2x - x = 10 ⇒ x = 10
∴ Area of trapezium = 10 × 3 = 30cm2
A person buys tea of three different qualities at Rs. 800, Rs. 500, and Rs. 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at Rs. 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is
Let the quantity bought are 6kg, 9kg & 15kg respectively. Total cost = 6 × 800 + 500 × 9 + 300 × 15 = Rs 13800 Total SP = 13800 × 1.5 = Rs 20700 SP of 5kg = 5 × 700 = Rs 3500 Remaining SP = Rs 17200 ∴ SP/kg = 17200/ 25= Rs. 688
For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4- digit number satisfying the above conditions is (in numerical value)
Let the number is abcd Given the a + b + c = 14 -----(1) b + c + d = 15 -----(2) & c = d + 4 -----(3) (2) - (1) ⇒ d - a = 1 ⇒ a = d - 1 -----(4) As highest number is asked ⇒ a should be max. So (4) ⇒ d should be maximum, or (3) ⇒ c should be maximum ⇒ c = 9 ⇒ d = 5 ∴ (4) ⇒ a = 4 & (1) ⇒ 4 + b + 9 = 14 ⇒ b = 1 ∴ Number is 4195.
The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is (in numerical value)
We have 15 identical balloons, 6 identical pencils and 3 identical erasers. Let the children are A, B and C. Now every child should get four balloons and one pencil. So first distribute 4 balloons and one pencil to A, B and C. We are left with 3 balloons, 3 pencils and 3 erasers. 3 balloons can be distributed to A, B, C in 3 + 3- 1C3 – 1 Or 5C2 = 10 ways. Similarly, 3 pencils can & 3 erasers can be distributed in 10 ways each. ∴ Total ways = 10 × 10 × 10 = 1000
For all possible integers n satisfying 2.25 ≤ 2 + 2n + 2 ≤ 202, the number of integer values of 3 + 3n + 1 is (in numerical value)
Given that 2.25 ≤ 2 + 2n+2 ≤ 202 ⇒ 2 + 0.25 ≤ 2 + 2n+2 ≤ 2 + 128 < 202 ⇒ 2 + 2-2 ≤ 2 + 2n+2 ≤ 2 + 27 ∴ - 2 ≤ n + 2 ≤ 7 ⇒ - 4 ≤ n ≤ 5 ∴ n = - 4, -3, -2, -1, 0, 1, 2, 3, 4, 5 Now for n = -1, 0, 1, 2, 3, 4, 5, 3 + 3n+1 is an integer. Therefore total ‘7’ values of ‘n’ are possible.
Raj invested Rs. 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than Rs. 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is
Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is
In the first case A worked for 8 hours and B worked for 7 hours. In the second case A worked for 4 hours and B worked for 2 hours. ∴ If B worked 5 hours less then time saved for A = 4 hours B does not work then time saved for A = 4/5 × 7 = 5.6 hours So in this A would have filled the tank in 8 - 5.6 = 2.4 hours or 144 minutes.
A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is (in numerical value)
Let the metallic white and metallic black balls are x each. Now 40% of white balls = x ⇒ white balls = 2.5x & 50% of black balls = x ⇒ Total black balls = 2x Now 2.5x + 2x = 450 ⇒ 4.5x = 450 ⇒ x = 100 ∴ Number of non-metallic white & non metallic black balls = 1.5x + x = 2.5x = 2.5 × 100 = 250
From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is (in numerical value)
In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is (in numerical value)
Let the original matches be ‘n’ and original goals be ‘x’ Now (n + 10) × 0.15 = x + 1 ⇒ 0.15 n + 1.5 = x + 1 ---(1) & (n + 10) × 0.2 = x + 2 ⇒ 0.2n + 2 = x + 2 ---(2) (2) – (1) ⇒ 0.05n + 0.5 = 1 ⇒ 0.05n = 0.5 ⇒ n = 0.5/ 0.05= 10
Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of Rs. 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is
We at Quantifiers understand and deliver on the personal attention each of our students requires. Whether it is through our pedagogy that enables non-engineers or non-math background students, our constant effort to proactively provide solutions, or our focus on our student’s goals.
